Product Series Formula through Induction For the series:
$(1+\frac{1}{1})^1$$(1+\frac{1}{2})^2$$.....$ $(1+\frac{1}{n})^n$ I have the formula  $\frac{(n+1)^n}{n!}$ for n$\in$ $\Bbb N$
I used induction to try and solve but I'm stuck at trying to prove it for n+1 since
$(1+\frac{1}{1})^1$$(1+\frac{1}{2})^2$$.....$$(1+\frac{1}{n})^n$$(1+\frac{1}{n+1})^{n+1}$ =  $\frac{(n+1)^n}{n!}$$(1+\frac{1}{1})^{n+1}$ = $\frac{(n+1)^n}{n!}$$\sum_{k=0}^n \binom{n}{k}(\frac{1}{n+1})^k$ using the binomial theorem
but I can't tell properly how to get  $\frac{(n+2)^{n+1}}{(n+1)!}$
Any hints? because I swear I'm not seeing something.
Thank you!

Hint: We have
\begin{eqnarray*}
\left( 1+\frac{1}{1} \right) \left( 1+\frac{1}{2} \right)^2 \cdots \left( 1+\frac{1}{n} \right)^n = \frac{(n+1)^{n}}{n!} 
\end{eqnarray*}
So
\begin{eqnarray*}
\left( 1+\frac{1}{1} \right) \left( 1+\frac{1}{2} \right)^2 \cdots \left( 1+\frac{1}{n} \right)^n \color{red}{\left( 1+\frac{1}{n+1} \right)^{n+1}}&=& \frac{(n+1)^{n}}{n!} \color{red}{\left(\frac{n+2}{n+1} \right)^{n+1}}\\
&=& \cdots
\end{eqnarray*}
