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.