Binomial identity

\displaystyle \sum_{k=1}^n(-1)^{n-k}\binom{n}{k}k^j=\begin{cases} n!, & j=n\\ 0, & j=1,2,\ldots,n\end{cases}
Proof. Consider the number of surjective functions f:\{1,2,\ldots,j\}\to\{1,2,\ldots,n\}. If we count them using the inclusion-exclusion principle we get the sum in the problem. On the other hand this number is trivially n! if j=n and 0 if j<n.
Remark 1. For j>n we again find the number of surjections, which has a closed form form some j.
Remark 2. Interesting identity which represents the above identities in a unified manner is the following \sum_{k=0}^n(-1)^{n-k}\binom{n}{k}(x+k)^n=n! for all x (in \mathbb{R} for example, but any commutative ring suffices).

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