3.1 The proof of Theorem 1.1
Suppose that (1.1) admits a nontrivial meromorphic solution f(z) such that \rho (f)<k+1, then by Lemma 2.1, for any given ϵ such that 0<\u03f5<min\{k+1\rho ,k\alpha ,\frac{{\beta}_{0}{\alpha}_{0}}{{\beta}_{0}+{\alpha}_{0}}\}, we have
exp\{{r}^{\rho (f)1+\u03f5}\}\le \left\frac{f(z+j)}{f(z)}\right\le exp\left\{{r}^{\rho (f)1+\u03f5}\right\},\phantom{\rule{1em}{0ex}}j=0,\dots ,n,
(3.1)
for all r outside of a possible exceptional set {E}_{1} with finite logarithmic measure.
Applying Lemma 2.2, we have
exp\{{r}^{\alpha +\u03f5}\}\le {A}_{j}(z)\le exp\left\{{r}^{\alpha +\u03f5}\right\},\phantom{\rule{1em}{0ex}}j=0,\dots ,n,
(3.2)
exp\{{r}^{\alpha +\u03f5}\}\le {B}_{0}(z)\le exp\left\{{r}^{\alpha +\u03f5}\right\},
(3.3)
for all r outside of a possible exceptional set {E}_{2} with finite linear measure.
Therefore from (1.1) we can get that
\frac{{A}_{n}(z)f(z+n)}{f(z)}+\cdots +\frac{{A}_{2}(z)f(z+2)}{f(z)}+\frac{{A}_{1}(z){e}^{{\alpha}_{1}{z}^{k}}f(z+1)}{f(z)}+{A}_{0}(z){e}^{{\alpha}_{0}{z}^{k}}+{B}_{0}(z){e}^{{\beta}_{0}{z}^{k}}=0,
i.e.,
\begin{array}{r}{A}_{0}(z){e}^{{\alpha}_{0}{z}^{k}}+{B}_{0}(z){e}^{{\beta}_{0}{z}^{k}}\\ \phantom{\rule{1em}{0ex}}\le \left\frac{{A}_{n}(z)f(z+n)}{f(z)}\right+\cdots +\left\frac{{A}_{2}(z)f(z+2)}{f(z)}\right+\left\frac{{A}_{1}(z){e}^{{\alpha}_{1}{z}^{k}}f(z+1)}{f(z)}\right.\end{array}
(3.4)
Setting {\alpha}_{0}={\alpha}_{0}{e}^{i{\theta}_{0}}, {\beta}_{0}={\beta}_{0}{e}^{i{\phi}_{0}} ({\theta}_{0},{\phi}_{0}\in [\frac{\pi}{2},\frac{3\pi}{2})).
Case 1. arg{\alpha}_{0}\ne \pi, which is {\theta}_{0}\ne \pi.
Subcase 1.1. Assume that {\theta}_{0}\ne {\phi}_{0}. By Lemma 2.4, for the above ϵ, there is a ray argz=\theta such that \theta \in [\frac{\pi}{2k},\frac{\pi}{2k})\mathrm{\setminus}({E}_{1}\cup {E}_{2}\cup {E}_{5}\cup {E}_{6}) (where {E}_{5} and {E}_{6} are defined as in Lemma 2.4, {E}_{1}\cup {E}_{2}\cup {E}_{5}\cup {E}_{6} is of linear measure zero) satisfying \delta ({\alpha}_{0}{z}^{k},\theta )>0, \delta ({\beta}_{0}{z}^{k},\theta )<0 or \delta ({\alpha}_{0}{z}^{k},\theta )<0, \delta ({\beta}_{0}{z}^{k},\theta )>0 for a sufficiently large r.
Since {A}_{j}(z), {B}_{0}(z) (≢0) (j=0,1,\dots ,n) are entire functions and max\{\rho ({A}_{j}),\rho ({B}_{0}):0\le j\le n\}=\alpha <k, then when \delta ({\alpha}_{0}{z}^{k},\theta )>0, \delta ({\beta}_{0}{z}^{k},\theta )<0 for a sufficiently large r, by Lemma 2.3, we have
{A}_{0}(z){e}^{{\alpha}_{0}{z}^{k}}>exp\{(1\u03f5)\delta ({\alpha}_{0}{z}^{k},\theta ){r}^{k}\},
(3.5)
{B}_{0}(z){e}^{{\beta}_{0}{z}^{k}}<exp\{(1\u03f5)\delta ({\beta}_{0}{z}^{k},\theta ){r}^{k}\}<1.
(3.6)
By (3.5) and (3.6), we have
\begin{array}{rcl}{A}_{0}(z){e}^{{\alpha}_{0}{z}^{k}}+{B}_{0}(z){e}^{{\beta}_{0}{z}^{k}}& \ge & {A}_{0}(z){e}^{{\alpha}_{0}{z}^{k}}{B}_{0}(z){e}^{{\beta}_{0}{z}^{k}}\\ >& exp\{(1\u03f5)\delta ({\alpha}_{0}{z}^{k},\theta ){r}^{k}\}1\\ =& (1o(1))exp\{(1\epsilon )\delta ({\alpha}_{0}{z}^{k},\theta ){r}^{k}\}.\end{array}
(3.7)
Since \theta \in [\frac{\pi}{2k},\frac{\pi}{2k})\mathrm{\setminus}({E}_{1}\cup {E}_{2}\cup {E}_{5}\cup {E}_{6}), we know that cosk\theta >0, then {e}^{{\alpha}_{1}{z}^{k}}={e}^{{\alpha}_{1}{r}^{k}cosk\theta}<1. Therefore, by (3.2) we obtain
{A}_{1}(z){e}^{{\alpha}_{1}{z}^{k}}\le exp\left\{{r}^{\alpha +\u03f5}\right\}.
(3.8)
Substituting (3.1), (3.2), (3.3), (3.7) and (3.8) into (3.4), we obtain
(1o(1))exp\{(1\u03f5)\delta ({\alpha}_{0}{z}^{k},\theta ){r}^{k}\}<nexp\{{r}^{\alpha +\u03f5}+{r}^{\rho (f)1+\u03f5}\}.
(3.9)
By \delta ({\alpha}_{0}{z}^{k},\theta )>0, \alpha +\u03f5<k and \rho (f)1+\u03f5<k, we know that (3.9) is a contradiction.
When \delta ({\alpha}_{0}{z}^{k},\theta )<0, \delta ({\beta}_{0}{z}^{k},\theta )>0, using a proof similar to the above, we can get a contradiction.
Subcase 1.2. Assume that {\theta}_{0}={\phi}_{0}. By Lemma 2.4, for the above ϵ, there is a ray argz=\theta such that \theta \in [\frac{\pi}{2k},\frac{\pi}{2k})\mathrm{\setminus}({E}_{1}\cup {E}_{2}\cup {E}_{5}\cup {E}_{6}) (where {E}_{5} and {E}_{6} are defined as in Lemma 2.4, {E}_{1}\cup {E}_{2}\cup {E}_{5}\cup {E}_{6} is of linear measure zero) satisfying \delta ({\alpha}_{0}{z}^{k},\theta )>0.
Since {\alpha}_{0}\le {\beta}_{0}, {\alpha}_{0}\ne {\beta}_{0}, and {\theta}_{0}={\phi}_{0}, then {\alpha}_{0}<{\beta}_{0}, thus \delta ({\beta}_{0}{z}^{k},\theta )>\delta ({\alpha}_{0}{z}^{k},\theta )>0. For a sufficiently large r, by Lemma 2.3, we get
{A}_{0}(z){e}^{{\alpha}_{0}{z}^{k}}<exp\{(1+\u03f5)\delta ({\alpha}_{0}{z}^{k},\theta ){r}^{k}\},
(3.10)
{B}_{0}(z){e}^{{\beta}_{0}{z}^{k}}>exp\{(1\u03f5)\delta ({\beta}_{0}{z}^{k},\theta ){r}^{k}\}.
(3.11)
By (3.10) and (3.11), we get
\begin{array}{rcl}{A}_{0}(z){e}^{{\alpha}_{0}{z}^{k}}+{B}_{0}(z){e}^{{\beta}_{0}{z}^{k}}& \ge & {B}_{0}(z){e}^{{\beta}_{0}{z}^{k}}{A}_{0}(z){e}^{{\alpha}_{0}{z}^{k}}\\ >& exp\{(1\u03f5)\delta ({\beta}_{0}{z}^{k},\theta ){r}^{k}\}exp\{(1+\u03f5)\delta ({\alpha}_{0}{z}^{k},\theta ){r}^{k}\}\\ =& {M}_{1}exp\{(1+\u03f5)\delta ({\alpha}_{0}{z}^{k},\theta ){r}^{k}\},\end{array}
(3.12)
where {M}_{1}=exp\{[(1\u03f5)\delta ({\beta}_{0}{z}^{k},\theta )(1+\u03f5)\delta ({\alpha}_{0}{z}^{k},\theta )]{r}^{k}\}1.
Since 0<\u03f5<min\{k+1\rho ,k\alpha ,\frac{{\beta}_{0}{\alpha}_{0}}{{\beta}_{0}+{\alpha}_{0}}\}, we see that (1\u03f5)\delta ({\beta}_{0}{z}^{k},\theta )(1+\u03f5)\delta ({\alpha}_{0}{z}^{k},\theta )>0, then exp\{[(1\u03f5)\delta ({\beta}_{0}{z}^{k},\theta )(1+\u03f5)\delta ({\alpha}_{0}{z}^{k},\theta )]{r}^{k}\}>1, {M}_{1}>0.
Since \theta \in [\frac{\pi}{2k},\frac{\pi}{2k})\mathrm{\setminus}({E}_{1}\cup {E}_{2}\cup {E}_{5}\cup {E}_{6}), we know that cosk\theta >0, then {e}^{{\alpha}_{1}{z}^{k}}={e}^{{\alpha}_{1}{r}^{k}cosk\theta}<1. Therefore, by (3.2) we obtain
{A}_{1}(z){e}^{{\alpha}_{1}{z}^{k}}\le exp\left\{{r}^{\alpha +\u03f5}\right\}.
(3.13)
Substituting (3.1), (3.2), (3.3), (3.12) and (3.13) into (3.3), we obtain
{M}_{1}exp\{(1+\u03f5)\delta ({\alpha}_{0}{z}^{k},\theta ){r}^{k}\}\le nexp\{{r}^{\alpha +\u03f5}+{r}^{\rho (f)1+\u03f5}\}.
(3.14)
By \delta ({\alpha}_{0}{z}^{k},\theta )>0, \alpha +\u03f5<k and \rho (f)1+\u03f5<k, we know that (3.14) is a contradiction.
Case 2. {\alpha}_{0}<{\alpha}_{1}, which is {\theta}_{0}=\pi.
Subcase 2.1. Assume that {\theta}_{0}\ne {\phi}_{0}, then {\phi}_{0}\ne \pi. By Lemma 2.4, for the above ϵ, there is a ray argz=\theta such that \theta \in [\frac{\pi}{2k},\frac{\pi}{2k})\mathrm{\setminus}({E}_{1}\cup {E}_{2}\cup {E}_{5}\cup {E}_{6}) (where {E}_{5} and {E}_{6} are defined as in Lemma 2.4, {E}_{1}\cup {E}_{2}\cup {E}_{5}\cup {E}_{6} is of linear measure zero) satisfying \delta ({\beta}_{0}{z}^{k},\theta )>0. Since cosk\theta >0, we have \delta ({\alpha}_{0}{z}^{k},\theta )={\alpha}_{0}cos({\theta}_{0}+k\theta )={\alpha}_{0}cosk\theta <0. For a sufficiently large r, by Lemma 2.3, we get
{A}_{0}(z){e}^{{\alpha}_{0}{z}^{k}}<exp\{(1\u03f5)\delta ({\alpha}_{0}{z}^{k},\theta ){r}^{k}\}<1,
(3.15)
{B}_{0}(z){e}^{{\beta}_{0}{z}^{k}}>exp\{(1\u03f5)\delta ({\beta}_{0}{z}^{k},\theta ){r}^{k}\}.
(3.16)
By (3.15) and (3.16), we get
\begin{array}{rcl}{A}_{0}(z){e}^{{\alpha}_{0}{z}^{k}}+{B}_{0}(z){e}^{{\beta}_{0}{z}^{k}}& \ge & {B}_{0}(z){e}^{{\beta}_{0}{z}^{k}}{A}_{0}(z){e}^{{\alpha}_{0}{z}^{k}}\\ >& exp\{(1\u03f5)\delta ({\beta}_{0}{z}^{k},\theta ){r}^{k}\}1.\end{array}
(3.17)
Using the same reasoning as in Subcase 1.1, we can get a contradiction.
Subcase 2.2. Assume that {\theta}_{0}={\phi}_{0}, then {\theta}_{0}={\phi}_{0}=\pi. By Lemma 2.4, for the above ϵ, there is a ray argz=\theta such that \theta \in [\frac{\pi}{2k},\frac{3\pi}{2k})\mathrm{\setminus}({E}_{1}\cup {E}_{2}\cup {E}_{5}\cup {E}_{6}) (where {E}_{5} and {E}_{6} are defined as in Lemma 2.4, {E}_{1}\cup {E}_{2}\cup {E}_{5}\cup {E}_{6} is of linear measure zero), then cosk\theta <0, \delta ({\alpha}_{0}{z}^{k},\theta )={\alpha}_{0}cos({\theta}_{0}+k\theta )={\alpha}_{0}cosk\theta >0, \delta ({\beta}_{0}{z}^{k},\theta )={\beta}_{0}cos({\phi}_{0}+k\theta )={\beta}_{0}cosk\theta >0.
Since {\alpha}_{0}\le {\beta}_{0}, {\alpha}_{0}\ne {\beta}_{0}, and {\theta}_{0}={\phi}_{0}, then {\alpha}_{0}<{\beta}_{0}, thus \delta ({\beta}_{0}{z}^{k},\theta )>\delta ({\alpha}_{0}{z}^{k},\theta )>0. For a sufficiently large r, we get (3.10), (3.11) and (3.12) hold.
Using the same reasoning as in Subcase 1.2, we can get a contradiction. Thus we have \rho (f)\ge k+1.
3.2 The proof of Theorem 1.2
Since {A}_{j}(z), {B}_{0}(z) (≢0) (j=0,1,\dots ,n) are meromorphic functions and max\{\rho ({A}_{j}),\rho ({B}_{0}):0\le j\le n\}=\alpha <1, then when \delta ({\alpha}_{0}z,\theta )>0, \beta ({a}_{0}z,\theta )<0 for a sufficiently large r, by Lemma 2.5, we have
{A}_{0}(z){e}^{{\alpha}_{0}z}\ge exp\{(1\u03f5)\delta ({\alpha}_{0}z,\theta )r\},
(3.18)
{B}_{0}(z){e}^{{\beta}_{0}z}\le exp\{(1\u03f5)\delta ({\beta}_{0}z,\theta )r\}<1.
(3.19)
Then, using a similar argument to that of Theorem 1.1, we obtain a contradiction.
3.3 The proof of Theorems 1.3 and 1.4
Suppose that f(z) is a nonconstant meromorphic solution of (1.1) such that \rho (f)<\mathrm{\infty}. We first prove that \lambda (fz)=\rho (f). Submitting f(z)=g(z)+z into (1.1), we get
\begin{array}{r}{A}_{n}(z)g(z+n)+\cdots +{A}_{2}(z)g(z+2)+{A}_{1}(z){e}^{{\alpha}_{1}{z}^{k}}g(z+1)\\ \phantom{\rule{1em}{0ex}}+({A}_{0}(z){e}^{{\alpha}_{0}{z}^{k}}+{B}_{0}(z){e}^{{\beta}_{0}{z}^{k}})g(z)=D(z),\end{array}
where
\begin{array}{rcl}D(z)& =& z\{{A}_{n}(z)+\cdots +{A}_{2}(z)+{A}_{1}(z){e}^{{\alpha}_{1}{z}^{k}}+({A}_{0}(z){e}^{{\alpha}_{0}{z}^{k}}+{B}_{0}(z){e}^{{\beta}_{0}{z}^{k}})\}\\ \{n{A}_{n}(z)+\cdots +2{A}_{2}(z)+{A}_{1}(z){e}^{{\alpha}_{1}{z}^{k}}\}\not\equiv 0.\end{array}
Since max\{\rho ({A}_{j}),\rho ({B}_{0}):0\le j\le n\}=\alpha <k, we have \rho (D)\le k.
Now, for any given \u03f5>0, applying Lemma 2.1 and Theorem 1.1, we can deduce that
\begin{array}{r}m(r,\frac{1}{g(z)})\\ \phantom{\rule{1em}{0ex}}=m(r,\frac{{A}_{n}(z)g(z+n)+\cdots +{A}_{2}(z)g(z+2)+{A}_{1}(z){e}^{{\alpha}_{1}{z}^{k}}g(z+1)+({A}_{0}(z){e}^{{\alpha}_{0}{z}^{k}}+{B}_{0}(z){e}^{{\beta}_{0}{z}^{k}})g(z)}{D(z)g(z)})\\ \phantom{\rule{1em}{0ex}}\le \sum _{j=0}^{n}m(r,\frac{g(z+j)}{g(z)})+\sum _{j=0}^{n}T(r,{A}_{j}(z))+T(r,{e}^{{\alpha}_{1}{z}^{k}})+T(r,{B}_{0}(z))\\ \phantom{\rule{2em}{0ex}}+T(r,{e}^{{\alpha}_{0}{z}^{k}})+T(r,{e}^{{\beta}_{0}{z}^{k}})+T(r,D(z))+O(logr)\\ \phantom{\rule{1em}{0ex}}=O\left({r}^{\rho 1+\u03f5}\right)=S(r,g).\end{array}
This implies that
N(r,\frac{1}{fz})=N(r,\frac{1}{g})=T(r,g)+S(r,g)=T(r,f)+S(r,f),
then \lambda (fz)=\rho (f) follows.
Next, we assert that either k+1\le \rho (f)\le max\{\lambda (f),\lambda (\frac{1}{f})\}+1 or \rho (f)=k+1. If the assertion does not hold, we have max\{k,\lambda (f),\lambda (\frac{1}{f})\}+1<\rho (f)<\mathrm{\infty}.
Assume that z=0 is a zero (or a pole) of f(z) of order m. Applying the Hadamard factorization of a meromorphic function, we write f(z) as follows:
f(z)={z}^{m}\frac{{P}_{1}(z)}{{P}_{2}(z)}{e}^{Q(z)},
where {P}_{1}(z), {P}_{1}(z) are entire functions such that \rho ({P}_{1})=\lambda (f), \rho ({P}_{2})=\lambda (\frac{1}{f}) and Q(z) is a polynomial such that deg Q(z)=q>max\{k,\lambda (f),\lambda (\frac{1}{f})\}+1.
Now, we obtain from (1.1) that
\sum _{j=0}^{n}{h}_{j}(z){e}^{Q(z+j)}=0,
(3.20)
where
{h}_{0}=({A}_{0}(z){e}^{{\alpha}_{0}{z}^{k}}+{B}_{0}(z){e}^{{\beta}_{0}{z}^{k}}){z}^{m}\frac{{P}_{1}(z)}{{P}_{2}(z)},\phantom{\rule{2em}{0ex}}{h}_{1}={A}_{1}(z){e}^{{\alpha}_{1}{z}^{k}}{(z+1)}^{m}\frac{{P}_{1}(z+1)}{{P}_{2}(z+1)},
and
{h}_{j}(z)={A}_{j}(z){(z+j)}^{m}\frac{{P}_{1}(z+j)}{{P}_{2}(z+j)}\phantom{\rule{1em}{0ex}}(2\le j\le n+1).
Notice that deg(Q(z+h)Q(z+l))=q1>\rho ({h}_{j}) for 0\le h<l\le n and 0\le j\le n. Thus, Lemma 2.6 is valid for (3.20), hence we get that {h}_{j}(z)\equiv 0 for j=1,2,\dots ,n+1, a contradiction to our assumption. This completes our proof.
The proof of Theorem 1.4 is similar to that of Theorem 1.3.
3.4 The proof of Theorem 1.5
Let f (≢0) be a meromorphic solution of (1.4). Suppose that \rho (f)<2, then by Lemma 2.1, for any given \u03f5>0, there exists a subset E\subset (1,\mathrm{\infty}) of finite logarithmic measure such that for all z satisfying z=r\notin [0,1]\cup E, and as r sufficiently large, we have
\left\frac{f(z+j)}{f(z+l)}\right\le exp\left\{{r}^{\rho (f)1+\u03f5}\right\}\phantom{\rule{1em}{0ex}}(j=0,1,\dots ,n,j\ne l).
(3.21)
Set z=r{e}^{i\theta}, {\alpha}_{j}={\alpha}_{j}{e}^{i{\phi}_{j}} and \delta ({\alpha}_{j}z,\theta )={\alpha}_{j}cos({\phi}_{j}+\theta ) (j=0,1,\dots ,k). Then {E}_{1}=\{\theta \in [0,2\pi ):\delta ({\alpha}_{j}z,\theta )=0,j=0,1,\dots ,n\}\cup \{\theta \in [0,2\pi ):\delta ({\alpha}_{j}z{\alpha}_{i}z,\theta )=0,0\le i<j\le n\} is a set of linear measure zero.
Consider that {a}_{j}(z)={A}_{j}(z){e}^{{\alpha}_{j}z}, {A}_{j}(z) (≢0) are meromorphic functions and \rho ({A}_{j})=\alpha <1, by Lemma 2.5, for the above \u03f5>0, there exists a set {F}_{j}\in [0,2\pi ) of linear measure zero such that for any z=r{e}^{i\theta} satisfying \theta \in [0,2\pi )\mathrm{\setminus}({E}_{1}\cup {F}_{j}), and as r\to \mathrm{\infty}, we have

(i)
if \delta ({\alpha}_{j}z,\theta )>0, then
exp\{(1\u03f5)\delta ({\alpha}_{j}z,\theta )r\}\le \left{a}_{j}\left(r{e}^{i\theta}\right)\right\le exp\{(1+\u03f5)\delta ({\alpha}_{j}z,\theta )r\};
(3.22)

(ii)
if \delta ({\alpha}_{j}z,\theta )<0, then
exp\{(1+\u03f5)\delta ({\alpha}_{j}z,\theta )r\}\le \left{a}_{j}\left(r{e}^{i\theta}\right)\right\le exp\{(1\u03f5)\delta ({\alpha}_{j}z,\theta )r\}.
(3.23)
Set {E}_{2}={\bigcup}_{j=0}^{n}{F}_{j}, then {E}_{2} is a set of linear measure zero.
Since {\alpha}_{j} are distinct complex constants, then there exists only one l\in \{0,1,\dots ,n\} such that \delta ({\alpha}_{l}z,\theta )=max\{\delta ({\alpha}_{j}z,\theta ):j=0,1,\dots ,n\} for any \theta \in [0,2\pi )\mathrm{\setminus}({E}_{1}\cup {E}_{2}). Now we take a ray argz={\theta}_{0}\in [0,2\pi )\mathrm{\setminus}({E}_{1}\cup {E}_{2}) such that \delta ({\alpha}_{l}z,{\theta}_{0})>0.
Let {\delta}_{1}=\delta ({\alpha}_{l}z,{\theta}_{0}), {\delta}_{2}=max\{\delta ({\alpha}_{j}z,{\theta}_{0}):j=0,1,\dots ,l,j\ne l\}, then {\delta}_{1}>{\delta}_{2}. We discuss the following two cases.
Case 1. {\delta}_{2}>0. We rewrite (1.4) in the form
{a}_{l}(z)=\sum _{j=0}^{n}{a}_{j}(z)\frac{f(z+j)}{f(z+l)}\phantom{\rule{1em}{0ex}}(j\ne l).
(3.24)
By (3.21), (3.22) and (3.24), we get for z=r{e}^{i{\theta}_{0}} and sufficiently large r\notin [0,1]\cup E,
\begin{array}{rcl}exp\{(1\u03f5){\delta}_{1}r\}& \le & \left{a}_{l}\left(r{e}^{i{\theta}_{0}}\right)\right\le \sum _{j=0,j\ne l}^{n}exp\{(1+\u03f5)\delta ({\alpha}_{j}z,\theta )r\}exp\left\{{r}^{\rho (f)1+\u03f5}\right\}\\ \le & nexp\{(1+\u03f5){\delta}_{2}r\}exp\left\{{r}^{\rho (f)1+\u03f5}\right\}.\end{array}
(3.25)
When 0<2\u03f5<min\{\frac{{\delta}_{1}{\delta}_{2}}{{\delta}_{1}+{\delta}_{2}},2\rho (f)\}, by (3.25), we get
exp\left\{\frac{{\delta}_{1}{\delta}_{2}}{2}r\right\}\le nexp\left\{{r}^{\rho (f)1+\u03f5}\right\}.
This is impossible.
Case 2. {\delta}_{2}<0. By (3.21), (3.23) and (3.24), we get for z=r{e}^{i{\theta}_{0}} and sufficiently large r\notin [0,1]\cup E,
\begin{array}{rcl}exp\{(1\u03f5){\delta}_{1}r\}& \le & \left{a}_{l}\left(r{e}^{i{\theta}_{0}}\right)\right\le \sum _{j=0,j\ne l}^{n}exp\{(1\u03f5)\delta ({\alpha}_{j}z,\theta )r\}exp\left\{{r}^{\rho (f)1+\u03f5}\right\}\\ \le & nexp\left\{{r}^{\rho (f)1+\u03f5}\right\}.\end{array}
This is a contradiction. Hence we get \rho (f)\ge 2.
3.5 The proof of Theorem 1.6
Considering {a}_{j}(z)={A}_{j}(z){e}^{{\alpha}_{j}z}+{D}_{j}(z) (j=0,1,\dots ,n), {A}_{j}(z) (≢0), {D}_{j}(z) (≢0) are meromorphic functions and \rho ({A}_{j})=\alpha <1, \rho ({D}_{j})=\beta <1, by Lemmas 2.2 and 2.5, we know that for any given \u03f5>0, there exists a subset E\subset (1,\mathrm{\infty}) of finite logarithmic measure such that, for all z satisfying z=r\notin [0,1]\cup E, and as r\to \mathrm{\infty}, we have

(i)
if \delta ({\alpha}_{j}z,\theta )>0, then
\begin{array}{r}exp\{(1\u03f5)\delta ({\alpha}_{j}z,\theta )r\}exp\left\{{r}^{\beta +\u03f5}\right\}\\ \phantom{\rule{1em}{0ex}}<\left{a}_{j}\left(r{e}^{i\theta}\right)\right<exp\{(1+\u03f5)\delta ({\alpha}_{j}z,\theta )r\}+exp\left\{{r}^{\beta +\u03f5}\right\}\\ \phantom{\rule{1em}{0ex}}<(1+o(1))exp\{(1+\u03f5)\delta ({\alpha}_{j}z,\theta )r\};\end{array}
(3.26)

(ii)
if \delta ({\alpha}_{j}z,\theta )<0, then
\left{a}_{j}\left(r{e}^{i\theta}\right)\right<exp\{(1\u03f5)\delta ({\alpha}_{j}z,\theta )r\}+exp\left\{{r}^{\beta +\u03f5}\right\}<(1+o(1))exp\left\{{r}^{\beta +\u03f5}\right\}.
(3.27)
Then, using a similar argument to that of Theorem 1.5 and Theorem 1.1 and only replacing (3.22) (or (3.23)) by (3.26) (or (3.27)), we can prove Theorem 1.6.
3.6 The proof of Theorem 1.7
Let f(z)\not\equiv 0 be a meromorphic solution of (1.5). Suppose that \rho (f)<1, then by Lemma 2.7, we obtain \rho (F)=\rho ({\sum}_{j=0}^{k}{a}_{j}(z)f(z+j))=1. This contradicts \rho (F)<1, therefore we have \rho (f)\ge 1.
Suppose that there exist two distinct meromorphic solutions {f}_{1}\not\equiv 0, {f}_{2}\not\equiv 0 of Eq. (1.5) such that max\{\rho ({f}_{1}),\rho ({f}_{2})\}<2, then {f}_{1}{f}_{2} is a meromorphic solution of the homogeneous linear difference equation corresponding to (1.5) and \rho ({f}_{1}{f}_{2})<2. By Theorem 1.5 or Theorem 1.6, we get a contradiction. So Eq. (1.5) has at most one meromorphic solution {f}_{0} satisfying 1\le \rho ({f}_{0})\le 2.
Next we prove max\{\lambda ({f}_{0}),\lambda (\frac{1}{{f}_{0}})\}=\rho ({f}_{0}) in the case \rho ({f}_{0})=1. Suppose that max\{\lambda ({f}_{0}),\lambda (\frac{1}{{f}_{0}})\}<\rho ({f}_{0}), then by the Weierstrass factorization, we obtain
{f}_{0}(z)=\frac{{P}_{1}(z)}{{P}_{2}(z)}{e}^{Q(z)},
(3.28)
where {P}_{1}(z), {P}_{1}(z) are entire functions such that \rho ({P}_{1})=\lambda ({P}_{1})=\lambda ({f}_{0}), \rho ({P}_{2})=\lambda ({P}_{2})=\lambda (\frac{1}{{f}_{0}}) and Q(z) is a polynomial of degree 1.
In the case {a}_{j}(z)={A}_{j}(z){e}^{{\alpha}_{j}z}. Substituting (3.28) into (1.5), we get
\sum _{j=0}^{n}{A}_{j}(z)\frac{{P}_{1}(z+j)}{{P}_{2}(z+j)}{e}^{{\alpha}_{j}z+Q(z+j)}=F(z).
(3.29)
Since {\alpha}_{j} are distinct complex numbers, by Lemma 2.7, we obtain that the order of the lefthand side of (3.29) is 1. This contradicts \rho (F)<1.
For {a}_{j}(z)={A}_{j}(z){e}^{{\alpha}_{j}z}+{D}_{j}(z), by using an argument similar to the above, we also obtain a contradiction.
It is obvious that max\{\lambda ({f}_{0}),\lambda (\frac{1}{{f}_{0}})\}=\rho ({f}_{0}) provided that 1<\rho ({f}_{0})<2. Therefore we have max\{\lambda ({f}_{0}),\lambda (\frac{1}{{f}_{0}})\}=\rho ({f}_{0}).