Interesting combinatorial identity A friend of mine has shown me the following combinatorial identity:
$$
\sum_{k=0}^n{x\choose k}\frac{k\cdot k!}{x^{k+1}}
= 1 - {x\choose n+1}\frac{(n+1)!}{x^{n+1}}
$$
I feel like I lack tools to tackle that problem. Hints?

In this situation a simple induction on $n$ works. When $n=0$,
your formula becomes $0=1-1$, which is admittedly true. Now, if the formula
holds for $n$, we have
$$
\begin{array}{lcl}
\sum_{k=0}^{n+1}{x\choose k}\frac{k\cdot k!}{x^{k+1}}
&=& {x\choose n+1}\frac{(n+1)\cdot (n+1)!}{x^{n+2}}+\sum_{k=0}^{n}{x\choose k}\frac{k\cdot k!}{x^{k+1}} \\
&=&  {x\choose n+1}\frac{(n+1)\cdot (n+1)!}{x^{n+2}}+1 - {x\choose n+1}\frac{(n+1)!}{x^{n+1}} \\
&=&  1+{x\choose n+1}\Bigg(\frac{(n+1)\cdot (n+1)!}{x^{n+2}}-
\frac{(n+1)!}{x^{n+1}}\Bigg) \\
&=&  1+{x\choose n+1}\Bigg(\frac{(n+1)!}{x^{n+2}}(n+1-x)\Bigg) \\
&=&  1+{x\choose n+1}\frac{x-(n+1)}{n+2}\Bigg(\frac{(n+2)!}{x^{n+2}}\Bigg) \\
&=&  1-{x\choose n+2}\Bigg(\frac{(n+2)!}{x^{n+2}}\Bigg) \\
\end{array}
$$
so that the formula holds for $n+1$ also. This finishes the proof by induction.
