Euler's limit formula for Gamma

First assume $\mathrm{Re}z>0$. By [L2] and the functional equation inside [L1], $\frac{n!n^z}{z(z+1)\cdots (z+n)}=n^z{\int }_0^1{t}^{z-1}(1-t{)}^{n}dt={\int }_0^n{u}^{z-1}{(1-\frac{u}{n})}^{n}du\mathrm{.}$

For each fixed $u\ge 0$, $(1-u\mathrm{/}n{)}^n\to e^{-u}$, and on compact right-half-plane strips the integrands from step 1.1 are dominated by an integrable majorant. Therefore the integrals in step 1.1 converge locally uniformly to ${\int }_0^{\mathrm{\infty }}{u}^{z-1}{e}^{-u}du=\mathrm{\Gamma }(z)$. Hence the displayed limit formula holds on $\mathrm{Re}z>0$.

For $n\ge 1$, write $F_n(z):=\frac{n!n^z}{z(z+1)\cdots (z+n)}\mathrm{.}$ Let $K\subseteq \mathbb{C}\setminus \{0,-1,-2,\dots \}$ be compact. Choose $m\ge 0$ so that $K+m+1:=\{z+m+1:z\in K\}$ lies in the right half-plane. Repeatedly using $F_n(z)=\frac{z+n+1}{nz}{F}_n(z+1)$ gives $F_n(z)=\left({\prod }_{j=0}^m\frac{z+n+j+1}{n(z+j)}\right)F_n(z+m+1)\mathrm{.}$ On $K$, the prefactor converges uniformly to $1\mathrm{/}(z(z+1)\cdots (z+m))$, while step 2.1 applied on $K+m+1$ gives $F_n(z+m+1)\to \mathrm{\Gamma }(z+m+1)$ uniformly there. By [L1], $\mathrm{\Gamma }(z)=\frac{\mathrm{\Gamma }(z+m+1)}{z(z+1)\cdots (z+m)}$ on $K$, so $F_n\to \mathrm{\Gamma }$ uniformly on $K$. Since $K$ was arbitrary, the limit formula holds locally uniformly on the whole pole-free set.