TITLE: Proof of the summation $n!=\sum_{k=0}^n \binom{n}{k}(n-k+1)^n(-1)^k$?
QUESTION [8 upvotes]: $$n!=\sum_{k=0}^n \binom{n}{k}(n-k+1)^n(-1)^k$$
Could anyone give the proof of the above equation? Thanks in advance!

REPLY [3 votes]: Suppose we seek to evaluate
$$\sum_{k=0}^n {n\choose k} (-1)^k (n-k+1)^n.$$
Introduce
$$(n-k+1)^n = 
\frac{n!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
\exp((n-k+1)z) \; dz.$$
We get for the sum
$$\frac{n!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
\exp((n+1)z) 
\sum_{k=0}^n {n\choose k} (-1)^k \exp(-kz)
\; dz
\\ = \frac{n!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
\exp((n+1)z) 
(1-\exp(-z))^n
\; dz
\\ = \frac{n!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
\exp(z) 
(\exp(z)-1)^n
\; dz.$$
This is $$n! [z^n] \exp(z) (\exp(z)-1)^n.$$
Now $$\exp(z)-1 = z + \frac{z^2}{2} + \frac{z^3}{6} +\cdots$$
and hence
$$(\exp(z)-1)^n = z^n + \cdots.$$
Therefore the result is
$$n! [z^0] \exp(z) = n!.$$