Proof that $\prod_{k=1}^{n-1}(1+\frac1k)^k = \frac{n^n}{n!}$ for all $n \in \Bbb N \ge 2$ I've tried to prove this for a while now, but I can't get it:
$\prod_{k=1}^{n-1}(1+\frac1k)^k = \frac{n^n}{n!}$ for all $n \in \Bbb N \ge 2$
Solution:
$\prod_{k=1}^{(n+1)-1}(1+\frac1k)^k=\frac{(n+1)^{(n+1)}}{(n+1)!}$
$\left(1+\frac1n\right)^n\cdot\prod_{k=1}^{n-1}\left(\left(1+\frac1k\right)^k\right)= \frac{(n+1)^n\cdot(n+1)}{n!\cdot(n+1)}$
$\frac{n^n}{n!}\left(1+\frac1n\right)^n= \frac{(n+1)^n}{n!}$
$n^n(1+\frac{1}{n})^n=(n+1)^n$
$(n+1)^n=(n+1)^n$

First, we can write out terms of the product as 
$$\prod_{k=1}^{n-1}\left(1+\frac1k\right)^k=\left(\frac{2}{1}\right)^1\left(\frac{3}{2}\right)^2\left(\frac{4}{3}\right)^3\cdots \left(\frac{n-2}{n-3}\right)^{}\left(\frac{n-1}{n-2}\right)^{n-2}\left(\frac{n}{n-1}\right)^{n-1}$$
Then, note that the cancellation of numerator terms leave only the last term $n^{n-1}$, while partial cancellation of the denominator terms render the denominator $(1)\cdot (2)\cdot(3)\cdots (n-3)\cdot(n-2)\cdot(n-1)=(n-1)!$.
Putting it all together reveals
$$\begin{align}\prod_{k=1}^{n-1}\left(1+\frac1k\right)^k&=\frac{n^{n-1}}{(n-1)!}\\\\&=\frac{n^n}{n!}\end{align}$$
And we are done!
