Convergence of $\Gamma(z)=\lim_{n\to\infty}\frac{n^zn!}{\prod_{m=0}^n(z+m)}$ This question is a place to store proofs of the convergence of Euler's product formula for the gamma function:
$$\Gamma(z)=\lim_{n\to\infty}\frac{n^zn!}{\prod_{m=0}^n(z+m)}$$
which is convergent for all $z\in\Bbb C$ such that $-z\notin\Bbb N_0$. This is a self-answered question so that I can record this proof for future reference, but other (shorter?) proofs and simplification of my own proof are of course welcome.

For $\Re(z)>0$ this is usually done by comparing the two integrals:
$$I_1 = \Gamma(z) = \int_{0}^{+\infty}t^{z-1}e^{-t}\,dt,\qquad I_2=\int_{0}^{n}t^{z-1}\left(1-\frac{t}{n}\right)^n\,dt.$$
Integrating by parts $I_2$ we get the Euler's product, hence all the issues are in proving that
$$\int_{n}^{+\infty}t^{z-1}e^{-t}\,dt\quad\text{and}\quad \int_{0}^{n}t^{z-1}\left(e^{-t}-\left(1-\frac{t}{n}\right)^n\right)dt$$
converge to zero as long as $n\to +\infty$. This is usually done through the Bernoulli inequality.
