For all $n >0$, $ \left(1+\frac{1}{n} \right)^n = 1+ \sum_{k=1}^n\bigl[ \frac{1}{k!} \prod_{r=0}^{k-1}(1-\frac{r}{n}) \bigr]$ I am working through some problems on induction and I have been stuck on this one for a while. If anyone has any hints. I can show it is true for the $n=1$ and $n=2$ case but I am having difficulty on the induction step.
This is what I have so far,
$ (1+\frac{1}{n} )^n      =  1+ \sum\limits_{k=1}^n[ \frac{1}{k!}    \prod\limits_{r=0}^{k-1}(1-\frac{r}{n})  ]$
$n =1 $ case
$ (1+\frac{1}{1} )^1      = 1+  \sum\limits_{k=1}^1[ \frac{1}{k!}    \prod\limits_{r=0}^{k-1}(1-\frac{r}{1})  ]$
$ 2    =  1+  \frac{1}{1!}    \prod\limits_{r=0}^{0}(1-\frac{r}{1})  ]= 1+(1-\frac{0}{1})=1+1=2$   
So, 2=2. True
$n=2$ case
$ (1+\frac{1}{2} )^2      = 1+  \sum\limits_{k=1}^2[ \frac{1}{k!}    \prod\limits_{r=0}^{k-1}(1-\frac{r}{2})  ]$
$ (1+\frac{1}{2} )^2      = 1+ [ \frac{1}{1!}    \prod\limits_{r=0}^{0}(1-\frac{r}{2})  ]+\frac{1}{2!}    \prod\limits_{r=0}^{1}(1-\frac{r}{2})  ]$
$ (1+\frac{1}{2} )^2      = 1+ [ \frac{1}{1!}  (1-\frac{0}{2})  ]+[\frac{1}{2!}   (1-\frac{1}{2})  ]$
$ (1+\frac{1}{2} )^2      = 1+ [ \frac{1}{1!}  (1)  ]+[\frac{1}{2!}   (\frac{1}{2})  ]$
$ (1+\frac{1}{2} )^2      = 1+ 1+  \frac{1}{4}  ]$
And finally, $\frac{9}{4} = \frac{9}{4}$ 
So the n=2 step is also true.
Induction Step
Show that the $(n+1)^{th}$ case is true when the $n^{th}$ case is true.
Now, I assume the $n^{th}$ case is true. So,
$ (1+\frac{1}{n} )^n      =  1+ \sum\limits_{k=1}^n[ \frac{1}{k!}    \prod\limits_{r=0}^{k-1}(1-\frac{r}{n})  ]$
then I multiply both sides by $ (1+\frac{1}{n} )^{n+1} $, but I haven't gotten far from there. Any hints would be great!

s(n+1)
Hints:
1.(1 + 1/n+1)^n+1 = (1 + 1/n)^n(1+ 1/n)...
2.Split up the sums
3.Use the binomial Theorem
