TITLE: Find the sum $\sum _{ k=1 }^{ 100 }{ \frac { k\cdot k! }{ { 100 }^{ k } } } \binom{100}{k}$
QUESTION [9 upvotes]: Find the sum $$\sum _{ k=1 }^{ 100 }{ \frac { k\cdot k! }{ { 100 }^{ k } } } \binom{100}{k}$$
When I asked my teacher how can I solve this question he responded it is very hard, you can't solve it. I hope you can help me in solving and understanding the question.

REPLY [5 votes]: Using the notation $n^\underline{r}=\overbrace{n\ (n-1)\ (n-2)\cdots(n-r+1)}^{r\text{  terms}}$ for the  falling factorial , we have
$$\begin{align}
\sum_{k=1}^n\frac {k\cdot k!}{n^k}\binom nk&=
\sum_{k=1}^n\frac {\color{blue}k\cdot k!}{n^k}\cdot \frac {n^\underline{k}}{k!}\\
&=\sum_{k=1}^n\frac {n^\underline{k}}{n^k}\color{blue}{[n-(n-k)]}\\
&=n\underbrace{\sum_{k=1}^n\frac {n^\underline{k}}{n^k}-\frac{n^\underline{k+1}}{n^{k+1}}}_{\text{telescoping sum}}
&&\text{as   }n^\underline{k}(n-k)=n^\underline{k+1}\\
&=n 
&&\text{as  }n^{\underline{n+1}}=0
\end{align}$$
Putting $n=100$ gives
$$\sum_{k=1}^{100}\frac {k\cdot k!}{100^k}\binom {100}k=100\qquad\blacksquare$$