# Let $G(z)=\prod\limits_{n=1}^{\infty}\left(1+\frac z n\right)e^{-\frac z n}$. Show that it's an entire function and $G(z-1)=ze^{\gamma}G(z)$

Let $$G(z)=\prod_{n=1}^{\infty}\left(1+\frac z n\right)e^{-\frac z n}$$Show that it's an entire function and $$G(z-1)=ze^{\gamma}G(z)$$ where $$\gamma=\lim_{n\rightarrow \infty}\left(\sum_{k=1}^n\frac 1k-\log n\right)$$

I want to use sum convergence to show the infinite product convergence but here $$Re(1+\frac zn)e^{-\frac zn}>0$$ does not hold, I was stuck here and I assume it should coincides with the Weierstrass factorization of some entire function with zeros at $$-n$$, but the Weierstrass factorization take the form $$z^me^{g(z)}\Pi E_{p_n}(\frac z {a_n})$$ where $$g(z)$$ should be entire, but here I can't abstract such a function from $$e^{-\frac z n}$$.

I was preparing for some exam, and this comes from past-year papers.

Uniform convergence on every disk $|z|\le R$. Within this disk $z/n\to 0$ uniformly. Hence $$(1+z/n)e^{-z/n} = (1+z/n)(1-z/n + O(1/n^2)) = 1 + O(1/n^2)$$ with $O$ being uniform. Since $\sum 1/n^2$ converges, the product converges. In particular, $G$ is never zero apart from $z=-1,-2,-3,\dots$.
Functional equation. Let $G_N$ denote the product taken up to $n=N$. Since $$G_N(z) = \prod_{n=1}^{N} \frac{n+z}{n}\cdot \prod_{n=1}^N \exp\left(-\frac{z}{n}\right)$$ and $$G_N(z-1) =z \prod_{n=1}^{N-1} \frac{n+z}{n+1} \cdot \prod_{n=1}^N \exp\left(-\frac{z-1}{n}\right)$$ we have $$\frac{G_N(z-1)}{G_N(z)} = z \frac{N+z}{N} \,\prod_{n=1}^{N-1} \frac{n}{n+1} \cdot \prod_{n=1}^N \exp\left(\frac{1}{n} \right)$$ This simplifies to $$z \frac{N+z}{N} \frac{1}{N} \exp\left(\sum_{n=1}^N \frac{1}{n} \right)$$ which converges to $z e^\gamma$ as $N\to\infty$.