Find the sum $\sum _{ k=1 }^{ 100 }{ \frac { k\cdot k! }{ { 100 }^{ k } } } \binom{100}{k}$

\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}

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 $$\sum_{k=1}^{100}\frac {k\cdot k!}{100^k}\binom {100}k=100\qquad\blacksquare$$

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}