In this paper, we use the basic notations of Nevanlinna theory [1, 2]. Given a meromorphic function f(z), recall that \alpha (z)\not\equiv 0,\mathrm{\infty} is a small function with respect to f(z) if T(r,\alpha )=S(r,f), where S(r,f) is used to denote any quantity satisfying S(r,f)=o(T(r,f)), and r\to \mathrm{\infty} outside of a possible exceptional set of finite logarithmic measure.
A Borel exceptional polynomial of f(z) is any polynomial p(z) satisfying
\lambda (f(z)p(z))=\underset{r\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\frac{{log}^{+}N(r,\frac{1}{f(z)p(z)})}{logr}<\rho (f),
where \lambda (f(z)p(z)) is the exponent of the convergence of zeros of f(z)p(z) and \rho (f) is the order of f(z). In the following, we assume that c is a nonzero complex constant, n and k are positive integers unless otherwise specified.
The zero distribution of differential polynomials is a classical topic in the theory of meromorphic functions. Hayman [[3], Theorem 10] firstly considered the value distribution of {f}^{n}{f}^{\prime}1, where f is a transcendental function. Then later this topic was considered by several authors such as [4, 5].
Theorem A ([[5], Theorem 1])
Let f be a transcendental meromorphic function. Ifn\ge 1, then{f}^{n}{f}^{\prime}1has infinitely many zeros.
Since {f}^{n}{f}^{\prime} can be written as \frac{{({f}^{n+1})}^{\prime}}{n+1}, Wang and Fang [6] improved Theorem A by proving the following result.
Theorem B ([[6], Corollary 1])
Let f be a transcendental meromorphic function. Ifn\ge k+1, then{({f}^{n})}^{(k)}1has infinitely many zeros.
The difference logarithmic derivative lemma, given by Chiang and Feng [[7], Corollary 2.5], Halburd and Korhonen [8], Theorem 2.1], [9], Theorem 5.6], plays an important part in considering the difference analogues of Nevanlinna theory. With the development of difference analogue of Nevanlinna theory, many authors paid their attention to the zero distribution of difference polynomials [10–18]. Laine and Yang [12], Theorem 2] firstly considered the zero distribution of f{(z)}^{n}f(z+c)a, where a is a nonzero constant, Liu and Yang [[13], Theorems 1.2 and 1.4] also considered the zeros of f{(z)}^{n}f(z+c)p(z) and f{(z)}^{n}{\mathrm{\Delta}}_{c}fp(z), where {\mathrm{\Delta}}_{c}f:=f(z+c)f(z) and p(z) is a nonzero polynomial. These results are summarized in Theorem C below, and they can be seen as difference analogues of Theorem A.
Theorem C Let f be a transcendental entire function of finite order andp(z)be a nonzero polynomial. Ifn\ge 2, thenf{(z)}^{n}f(z+c)p(z)has infinitely many zeros. If f is not a periodic function with period c andn\ge 2, thenf{(z)}^{n}{\mathrm{\Delta}}_{c}fp(z)has infinitely many zeros.
As the results on the difference analogues of Theorem B, Liu, Liu and Cao [14] investigated the zeros of {[f{(z)}^{n}f(z+c)]}^{(k)}\alpha (z) and {[f{(z)}^{n}{\mathrm{\Delta}}_{c}f]}^{(k)}\alpha (z), where \alpha (z) is a nonzero small function with respect to f(z). Some results can also be found in [16] on the case that \alpha (z) is a nonzero polynomial.
Theorem D ([[14], Theorems 1.1 and 1.3])
Let f be a transcendental entire function of finite order and\alpha (z)be a nonzero small function with respect tof(z). Ifn\ge k+2, then{[f{(z)}^{n}f(z+c)]}^{(k)}\alpha (z)has infinitely many zeros. If f is not a periodic function with period c andn\ge k+3, then{[f{(z)}^{n}{\mathrm{\Delta}}_{c}f]}^{(k)}\alpha (z)has infinitely many zeros.
However, we remark that the zeros of {[f{(z)}^{n}f(z+c)]}^{(k)} or {[f{(z)}^{n}{\mathrm{\Delta}}_{c}f]}^{(k)} were not mentioned in [14, 16]. We will consider this problem in this paper, some ideas of proofs partially relying on the ideas used in [14, 16].
It is easy to know that if f has infinitely many zeros, then f{(z)}^{n}f(z+c) must have infinitely many zeros. In fact, we mainly get some results on the case that n=1. The following example shows that f(z)f(z+c) can admit infinitely many zeros, however, {[f(z)f(z+c)]}^{(k)} has finitely many zeros.
Example 1 Suppose that {e}^{c}=1 and f(z)={e}^{z}+1. Then f(z)f(z+c)=1{e}^{2z} has infinitely many zeros, but {[f(z)f(z+c)]}^{(k)}={2}^{k}{e}^{2z} has no zeros.
What conditions will guarantee that {[f(z)f(z+c)]}^{(k)} can admit infinitely many zeros? Obviously, the value 1 is the Borel exceptional value of f(z) in Example 1. Here, we obtain the following result.
Theorem 1.1 Let f be a finite order transcendental entire function with a Borel exceptional polynomialq(z). Then the following statements hold.

(i)
If q(z)\equiv 0, then {[f(z)f(z+c)]}^{(k)} has no nonzero Borel exceptional value.

(ii)
If q(z)\not\equiv 0 and deg(q(z))<\frac{k}{2}, then {[f(z)f(z+c)]}^{(k)} has infinitely many zeros, except in the case f(z)=Aq(z){e}^{\alpha z}+q(z) and {e}^{c}=1, where A is a nonzero constant.

(iii)
If deg(q(z))\ge \frac{k}{2}, then {[f(z)f(z+c)]}^{(k)} has infinitely many zeros.
Remark 1 (1) If q(z)\equiv 0, then {[f(z)f(z+c)]}^{(k)} can admit finitely or infinitely many zeros. For example, if f(z)=z{e}^{z} and {e}^{c}=2, then {[f(z)f(z+c)]}^{(k)}=p(z){e}^{2z} has finitely many zeros, where p(z) is a polynomial in z. If f(z)=({e}^{z}+1){e}^{{z}^{2}} and {e}^{c}=1, thus the value 0 is a Borel exceptional value of f(z), then {[f(z)f(z+c)]}^{\prime}=[(4z+2c)(5z+2c){e}^{2z}]{e}^{{z}^{2}+{(z+c)}^{2}} has infinitely many zeros, but \lambda ({[f(z)f(z+c)]}^{\prime})=1<\rho ({[f(z)f(z+c)]}^{\prime})=2.

(2)
From Example 1, we know that the exceptional case in (ii) can occur.
If f has finitely many zeros, then f{(z)}^{n}f(z+c) must have finitely many zeros; however, f{(z)}^{n}{\mathrm{\Delta}}_{c}f can admit infinitely many zeros. The following example shows that the zero distribution of f{(z)}^{n}{\mathrm{\Delta}}_{c}f is different from that of f{(z)}^{n}f(z+c).
Example 2 Suppose that {e}^{c}=1 and f(z)={e}^{{z}^{2}}. Then f{(z)}^{n}f(z+c) has no zeros, but f{(z)}^{n}{\mathrm{\Delta}}_{c}f={e}^{(n+1){z}^{2}}({e}^{2zc+{c}^{2}}1) has infinitely many zeros.
Chen [10] investigated the problem: what conditions will guarantee that f{(z)}^{n}{\mathrm{\Delta}}_{c}f have infinitely many zeros. From the following Example 3, we know that the zero distribution of {[f{(z)}^{n}{\mathrm{\Delta}}_{c}f]}^{(k)} may be different from that of f{(z)}^{n}{\mathrm{\Delta}}_{c}f.
Example 3 Suppose that f(z)={e}^{z}+z of \rho (f)=1 and {e}^{c}=1. Then f(z){\mathrm{\Delta}}_{c}f=c({e}^{z}+z) has infinitely many zeros, but {[f(z){\mathrm{\Delta}}_{c}f]}^{(k)}=c{e}^{z} has no zeros, where k\ge 2.
Thus, it is natural to consider what conditions can guarantee that {[f{(z)}^{n}{\mathrm{\Delta}}_{c}f]}^{(k)} have infinitely many zeros. We obtain the following theorems.
Theorem 1.2 Let f be a transcendental entire function with finite order, n\ge 2, {\mathrm{\Delta}}_{c}f\not\equiv 0. Iff(z)has finitely many zeros and\rho (f)\ne 1, then{[f{(z)}^{n}{\mathrm{\Delta}}_{c}f]}^{(k)}has infinitely many zeros. Iff(z)has finitely many zeros and\rho (f)=1, then{[f{(z)}^{n}{\mathrm{\Delta}}_{c}f]}^{(k)}has finitely many zeros.
In the case of n=1, by using a similar method as in the proof of Theorem 1.1, we have the following result.
Theorem 1.3 Let f be a finiteorder transcendental entire function with a Borel exceptional polynomialq(z), k be a positive integer, {\mathrm{\Delta}}_{c}f\not\equiv 0. Then the following statements hold.

(i)
If q(z)\equiv 0, then {[f(z){\mathrm{\Delta}}_{c}f]}^{(k)} has no nonzero Borel exceptional value.

(ii)
If q(z)\not\equiv 0 and degq(z)<\frac{k+1}{2}, then {[f(z){\mathrm{\Delta}}_{c}f]}^{(k)} has infinitely many zeros, except in the case f(z)=A{e}^{\alpha z}+q(z) and {e}^{c}=1, where q(z) is not a constant and A is a nonzero constant.

(iii)
If degq(z)\ge \frac{k+1}{2}, then {[f(z){\mathrm{\Delta}}_{c}f]}^{(k)} has infinitely many zeros.
Remark 2 From Example 3, we know that the exceptional case in (ii) can occur.