How do I prove $\lim_{n\to\infty} \frac{n^z n!}{z(z+1)\cdots (z+n)}=\frac{1}{z}\prod_{n=1}^\infty \frac{(1+\frac{1}{n})^z}{1+\frac{z}{n}}$? How do I prove $$\lim_{n\to\infty} \frac{n^z n!}{z(z+1)\cdots (z+n)}=\frac{1}{z}\prod_{n=1}^\infty \frac{(1+\frac{1}{n})^z}{1+\frac{z}{n}}$$?
This could be proven if $$\lim_{n\to\infty} \frac{\prod_{k=1}^n (1+\frac{1}{k})}{n} = 1$$ but I can neither prove this one.
Please help!
 A: Hint: a definition of $\Gamma $ is  $$\frac{1}{\Gamma(z)} = e^{\gamma z}\cdot z \cdot \prod_{n =1}^{\infty} (1+\frac{z}{n})e^{- \frac{z}{n}}$$  where $\gamma$ is the Euler constant
A: To prove that 
$\large \lim_{n \to \infty}\frac{\prod_{k=1}^n(1+\frac{1}{k})}{n}=1$
expand out the numerator, which will result in
$\prod_{k=1}^n(1+\frac{1}{k})=\prod_{k=1}^n(\frac{k+1}{k})$
writing this out explicitly we have
$\frac{2}{1}\times \frac{3}{2} \cdots \times \frac{n}{n-1} \times \frac{n+1}{n}$
notice that the numerator of the first fraction, all the intermediate fractions, and the denominator of the last fraction cancels out, leaving $n+1$
Thus 
$\large \lim_{n \to \infty}\frac{\prod_{k=1}^n(1+\frac{1}{k})}{n}=\lim_{n \to \infty}\frac{n+1}{n}\rightarrow 1$ 
A: Hint:
\begin{align}
\lim_{n\to\infty} \frac{n^z n!}{z(z+1)\cdots (z+n)} &= \frac{1}{z} \lim_{n \to \infty} \frac{ \left( \frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdots \frac{n}{n-1} \right)^z n!}{(1+\frac{z}{1})(1 + \frac{z}{2}) (1 + \frac{z}{3}) \cdots (1 + \frac{z}{n}) \cdot n! } \\
&= \frac{1}{z} \lim_{n \to \infty} \frac{ \left\{ \left( 1 + \frac{1}{1} \right) \cdot \left( 1 + \frac{1}{2} \right) \cdot \left( 1 + \frac{1}{3} \right) \cdots \left( 1 + \frac{1}{n} \right) \right\}^z}{(1+\frac{z}{1})(1 + \frac{z}{2}) (1 + \frac{z}{3}) \cdots (1 + \frac{z}{n}) } 
\end{align}
using that as $n \to \infty$, $\frac{n}{n-1}, \frac{n+1}{n} \to 1$.
