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.
 A: It is sufficient to prove convergence of the log-gamma function,
$$\log\Gamma(z)=\lim_{n\to\infty}F(n):=\lim_{n\to\infty}z\log n-\log z-\sum_{k=1}^n\log(z/k+1),$$
because $e^x$ is continuous and $e^{F(n)}=\frac{n^zn!}{\prod_{m=0}^n(z+m)}$ so that $\Gamma(z)=e^{\log \Gamma(z)}$. (Note that this expression is not actually the principal log applied to the gamma function - the branch cuts are different.)
The zeta series $\sum_{k=1}^\infty\frac1{k^2}$ is convergent and hence Cauchy, and so there is an $N\ge\max(2,2|z|)$ such that $\sum_{k=m'}^n\frac1{k^2}<\frac{\epsilon}{2(|z|+|z|^2)}$ for any $N\le m'\le n$. We will show that $F(n)$ is Cauchy. Now, the order $1$ taylor expansion for $\log$, with error term, is $$|\log(1+x)-x|\le\frac{|x|^2}{1-|x|}\qquad\mbox{for }|x|<1.$$
Thus for any $k\ge N$,
\begin{align}
\left|z\log\Big(1-\frac1k\Big)+\log\Big(\frac zk+1\Big)\right|&=
\left|z\bigg(\!\log\Big(1-\frac1k\Big)+\frac1k\!\bigg)+\bigg(\!\log\Big(1+\frac zk\Big)-\frac zk\!\bigg)\right|\\
&\le|z|\frac{(1/k)^2}{1-1/k}+\frac{(|z|/k)^2}{1-|z|/k}\\
&=\frac1{k^2}\!\left(|z|\frac k{k-1}+|z|^2\frac k{k-|z|}\right)\\
&\le\frac2{k^2}\!\left(|z|+|z|^2\right)
\end{align}
Each ratio in the last step is at most $2$ because $k\ge N\ge 2,2|z|$. Thus for any $N\le m<n$:
\begin{align}
|F(m)-F(n)|&=\left|\sum_{k=m+1}^n\bigg(z\log\Big(1-\frac1k\Big)+\log\Big(\frac zk+1\Big)\bigg)\right|\\
&\le\sum_{k=m+1}^n\frac2{k^2}\!\left(|z|+|z|^2\right)<\epsilon,
\end{align}
so $F$ is cauchy and $\Gamma$ is convergent.
A: 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.
