If $a$ is not an integer, show that $$f(z) =\prod_{n \in\Bbb Z\setminus\{0\}}\left(1+{z\over a+n}\right)e^{-z/(n+a)}$$ is an entire function.
Using the series expansion of $e^{-z/(n+a)}$, we get \begin{align*} \left(1+{z\over a+n}\right)e^{-z/(n+a)}& = \sum_{k=0}^\infty \left(1+{z\over a+n}\right){1\over k!}\left(-{z\over n+a}\right)^k\\ & = \sum_{k=0}^\infty \left(1+{z\over a+n}\right){1\over k!}(-1)^k{1\over (n+a)^k}z^k\\ & = \sum_{k=0}^\infty(-1)^k{1\over k!}\left({z^k\over(n+a)^k}+{z^{k+1}\over (a+n)^{k+1}}\right)\\ & = 1+\sum_{k=0}^\infty (-1)^k{k\over(k+1)!}{z^{k+1}\over (a+n)^{k+1}}.\\ \end{align*} Similary computation shows that \begin{align*} \left(1+{z\over a+n}\right)e^{-z/n} = 1+\sum_{k=1}^\infty (-1)^k{(a+n)-kn\over k!(a+n)n^k}z^k.\\ \end{align*} Now I'm trying to use the fact that if $a_n\neq -1$ for each $n$, $\sum_{n=1}^\infty a_n$ converges absolutely if and only if $\prod_{n=1}^\infty (1+|a_n|)$ converges. But I don't know how to show the convergence of infinite series in my case.
Note that you can make a more precise estimate:
$$ \left(1 + \frac{z}{a + n}\right)e^{-z/(n+a)} = \left(1 + \frac{z}{a + n}\right)\left(1 + -\frac{z}{a + n} + O(n^{-2})\right) = 1 + O(n^{-2}) $$
which converges; and in particular on every disc $|z| \leq R$, this convergence should be uniform; uniform convergence on compacts nets you holomorphicity.