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Let $$S_r(n)=\sum_{m=0}^n(-1)^m m^r\binom{n}{m}$$ Evaluate $S_n(n)$

For this question, the first part I did is to prove that for integer r, $0

I did this by induction, given the binomial coefficient $$(1+z)^n=\sum_{m=0}^nz^m \binom{n}{m}$$ and differentiate it $r$ times. The result can be obtained for $S_1(n)=S_2(n)=\dots=S_k(n)=0 \implies S_{k+1}=0, $ given $0

I want to show that $S_{n}(n)$ is a linear combination of all $S_k(n)$ for $0

How should I proceed?

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    If you just take the binomial expansion of $(1+z)^n$ and differentiate $r$ times, you won't get the sum you're looking for; you get $m(m-1)\cdots(m-r+1)$ instead of $m^r$. You have to differentiate-then-multiply-by-$z$ $r$ times. – 2017-01-16

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I managed to solve it later in the day and here is my solution, using the result that for all $0