TITLE: Using $\Gamma(z) = \lim_{n \to \infty} \left[ \frac{n! n^z}{z(z+1)(z+2)\ldots(z+n)} \right]$ to prove the Weiestrass product.
QUESTION [6 upvotes]: I was searching the web quite thoroughly in the last two days. I was in paralytically
looking for a rigorous proof using 
$$
   \Gamma(z) = \lim_{n \to \infty} \left[ \frac{n! n^z}{z(z+1)(z+2)\ldots(z+n)} \right]
$$
to prove the canonical wierstrass representation of the gamma function
$$
   \frac{1}{\Gamma(z)} = z e^{\gamma z} \prod_{n=1}^\infty \left( 1 + \frac{z}{n}\right) e^{-z/n}
$$
I have seen a proof for the reverse, example here
http://www.proofwiki.org/wiki/Equivalence_of_Gamma_Function_Definitions
but how does one go about proving it the other way?

REPLY [7 votes]: According to the limit,
$$
\begin{align}
\frac1{\Gamma(z)}
&=z\lim_{n\to\infty}n^{-z}\prod_{k=1}^n\left(1+\frac zk\right)\\
&=z\lim_{n\to\infty}n^{-z}\prod_{k=1}^ne^{z/k}\prod_{k=1}^n\left(1+\frac zk\right)e^{-z/k}\\
&=z\lim_{n\to\infty}e^{z(H_n-\log(n))}\prod_{k=1}^n\left(1+\frac zk\right)e^{-z/k}\\
&=ze^{\gamma z}\prod_{k=1}^\infty\left(1+\frac zk\right)e^{-z/k}\\
\end{align}
$$