Closed form of alternating binomial constant power series

Question

Is there any closed form of the sum $\sum_{k=0}^{n}\binom{n}{k}(-1)^{n-k}k^z$ for some fixed positive integer $z$.

Answer 1

$$\sum_{k=0}^{n} \binom{n}{k} (-1)^{n-k} k^{z}=0,$$ where $z$ is an integer, and $0 \le z < n$.

Answer 2

Notice that $$ x\frac{d}{dx} x^k = kx^k, $$ so you can find a closed form for any integer $z$ by applying the operator $x \frac{d}{dx}$ to both sides of $$ (x-1)^n = \sum_{k=0}^n \binom{n}{k} (-1)^{n-k}x^k $$ $z$ times, then setting $x=1$. For $0 \leq z < n$ you get Math Lover's answer because all the terms on the left contain at least one factor of $(1-x)$, which vanishes when you put $x=1$. For $z=n$, you get $n!$ because the only term remaining is when all of the derivatives are applied to $(x-1)^n$. (Both of these results were known to Euler: they are related to $n$th differences.) Beyond that, the formulae get quite messy.

See also Euler's formula nth Differences of Powers, H. W. Gould, The American Mathematical Monthly, Vol. 85, No. 6 (Jun. - Jul., 1978), pp. 450-467, for some history and other applications of this result, which I have cited on here before.