Proving $z\,\Gamma(z) = \Gamma(z+1)$ using the product formula I am trying to show that $z\,\Gamma(z) = \Gamma(z+1)$ using the product formula:
$$ \Gamma(z) = \frac{e^{-\gamma z}}{z} \prod\limits_{n=1}^\infty\left(1+\frac{z}{n}\right)^{-1}e^{z/n}$$
where $\gamma$ is the Euler-Mascheroni constant. 
The only thing I have been able to do is to take the natural logarithm of both sides to convert the product to a sum:
$$\ln(\Gamma(z+1))=-\gamma (z+1)-\ln(z+1)-\sum\limits_{n-1}^\infty \ln\left(1+\frac{z+1}{n}\right)+\sum\limits_{n-1}^\infty\frac{z+1}{n} $$
I have also made the observation that $\ln(z)+\ln\Gamma(z)=\ln \Gamma(z+1)$. I have no idea what to do afterwards, especially with the $\ln\left(1+\frac{z+1}{n}\right)$ term. 
 A: The key is when you divide $\Gamma(z+1)$ by $\Gamma(z)$, most pieces in the
infinite product of the quotient is forming a telescoping product. More precisely,
$$\begin{align}\frac{\Gamma(z+1)}{\Gamma(z)}
&= \frac{z e^{\gamma z}}{(z+1) e^{\gamma(z+1)}} 
\frac{
\prod\limits_{k=1}^\infty \left(1+\frac{z}{k}\right) e^{-z/k}
}{
\prod\limits_{k=1}^\infty \left(1+\frac{z+1}{k}\right) e^{-(z+1)/k}
}\\
&= \frac{z e^{-\gamma}}{z+1}\lim_{N\to\infty} \prod_{k=1}^N \frac{1+\frac{z}{k}}{1+\frac{z+1}{k}} e^{1/k}\\
&= \frac{z}{z+1}\lim_{N\to\infty} e^{H_N-\gamma} \prod_{k=1}^N \frac{z+k}{z+k+1}\\
&= z \lim_{N\to\infty} \frac{e^{H_N-\gamma}}{z+N+1}
\end{align}
$$
where $H_N = \sum_{k=1}^N \frac{1}{k}$ is the $N^{th}$ harmonic number. 
For large $N$, $H_N$ has the asymptotic expansion
$$H_N \sim \log N + \gamma + \frac{1}{2N} - \sum_{k=1}^\infty \frac{B_{2k}}{2kN^{2k}}$$
where $B_k$ are the Bernoulli numbers. As a result,
$$\frac{\Gamma(z+1)}{\Gamma(z)} = z \lim_{N\to\infty} \frac{N e^{O(1/N)}}{z + N + 1 } = z$$
A: $$\Gamma (x) =\frac{e^{-\gamma x}}{x}\prod_{i=1}^{\infty}\frac{e^{x/i}}{1+x/i}\\
x\Gamma(x)=e^{-\gamma (x+1)}.e^{\gamma}\prod_{i=1}^{\infty}\frac{e^{(x+1)/i}.e^{-1/i}}{1+x/i}=\Gamma(x+1)e^{\gamma}\prod_{i=1}^{\infty}\left[{e^{-1/i}}\frac{1+(x+1)/i}{1+x/i}\right]$$
Now
$$\lim_{n\to\infty}\prod_{i=1}^{n}e^{-1/i}\frac{1+(x+1)/i}{1+x/i}\\=\lim_{n\to\infty}\exp\left(-\left(\sum_{i=1}^{n}\frac1{i}\right)\right)\prod_{i=1}^{n}\frac{x+i+1}{x+i}\\\sim\lim_{n\to\infty}\exp(-\gamma-\log n)\frac{x+n+1}{x+1}=e^{-\gamma}$$
