Find the sum $\sum _{ k=1 }^{ 100 }{ \frac { k\cdot k! }{ { 100 }^{ k } } } \binom{100}{k}$ 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. 
 A: \begin{align}
\sum\limits_{k=1}^{100} \frac {k\cdot k!}{100^k} \frac{100!}{k!(100-k)!} &= \frac{100!}{100^{100}} \sum\limits_{k=1}^{100} \frac{k\cdot100^{100-k}}{(100-k)!}\\
&= \frac{100!}{100^{100}} \sum\limits_{k=0}^{99}\frac{(100-k)\cdot 100^k}{k!}\\ &=\frac{99!}{100^{99}} \sum\limits_{k=0}^{99}
\left( \frac {100^{k+1}}{k!} - \frac{100^k}{(k-1)!}\right)\\ &=
\frac{99!}{100^{99}} 
\left(\frac{100^1}{0!}+
\sum\limits_{k=1}^{99} \frac {100^{k+1}}{k!} -
\sum\limits_{k=0}^{98} \frac{100^{k+1}}{k!}\right)\\&=
\frac{99!}{100^{99}} 
\left(\frac{100^1}{0!}+
\frac{100^{100}}{99!} +
\sum\limits_{k=1}^{98} \frac {100^{k+1}}{k!} -
\sum\limits_{k=1}^{98} \frac{100^{k+1}}{k!}
-\frac{100^1}{0!}
\right)\\
&= 
100
\end{align}
A: 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
$$\sum_{k=1}^{100}\frac {k\cdot k!}{100^k}\binom {100}k=100\qquad\blacksquare$$
A: I am re-editing a-rodin's answer, correcting a few typos [of an earlier version, now edited].
\begin{align}
\sum\limits_{k=1}^{100} \frac {k\cdot k!}{100^k} \frac{100!}{k!(100-k)!} &= \frac{100!}{100^{100}} \sum\limits_{k=1}^{100} \frac{k\cdot100^{100-k}}{(100-k)!}\\
&= \frac{100!}{100^{100}} \sum\limits_{k=0}^{99}\frac{(100-k)\cdot 100^k}{k!}\\ &=\frac{100!}{100^{100}} \sum\limits_{k=1}^{99}
\left( \frac {100^{k+1}}{k!} - \frac{100^k}{(k-1)!}\right)+\frac{100!}{100^{99}}\\ &=
\frac{100!}{100^{100}} 
\left(
\sum\limits_{k=1}^{99} \frac {100^{k+1}}{k!} -
\sum\limits_{k=0}^{98} \frac{100^{k+1}}{k!}\right)+\frac{100!}{100^{99}}\\&=
\frac{100!}{100^{100}} 
\left(
\frac{100^{100}}{99!} +
\sum\limits_{k=1}^{98} \frac {100^{k+1}}{k!} -
\sum\limits_{k=1}^{98} \frac{100^{k+1}}{k!}
-\frac{100^1}{0!}
\right)+\frac{100!}{100^{99}}\\
&= 
100-\frac{100!}{100^{99}}+\frac{100!}{100^{99}}=100.
\end{align}
