0
$\begingroup$

Prove that

$$\sum_{k=0}^{n}\binom{n}{k}(-1)^kk^n = (-1)^nn!$$

Please can anyone help me out here?

$\endgroup$

marked as duplicate by Théophile, Dragonemperor42, N. F. Taussig, Claude Leibovici, JMP Mar 26 '17 at 8:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

0
$\begingroup$

Apply $\left(x \frac{d}{dx}\right)^n$ to both sides of the identity $(1-x)^n = \sum_{0 \leq k \leq n} \binom{n}{k} (-1)^k x^k $, then evaluate at $x=1$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.