Evaluate $S_n(n)$ for $S_r(n)=\sum_{m=0}^n(-1)^m m^r\binom{n}{m}$ 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, $0<r<n$, $S_r(n)=0$
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 $0<k<n$
I want to show that $S_{n}(n)$ is a linear combination of all $S_k(n)$ for $0<k<n$, but is seems not the case.
How should I proceed?

I managed to solve it later in the day and here is my solution, using the result that for all $0<r<n$ $$S_r(n)=0$$ 
So for $S_n(n)$ we differentiate $(1+z)^n$ n times,
$$n!=\sum_{m=0}^{n}\binom{n}{m}m(m-1)(m-2)\dots(m-n+1)z^{m-n}$$
Letting $z=-1$,
$$n!=\sum_{m=0}^{n}(-1)^m(-1)^n\binom{n}{m}m(m-1)(m-2)\dots(m-n+1)$$
Where the parts with $m$ are just linear combinations of $S_1(n),S_2(n),...S_{n-1}(n)=0$
So 
$$n!=\sum_{m=0}^{n}(-1)^m\binom{n}{m}[(-1)^nm^n+C_1m^{n-1}+C_2m^{n-2}+...]=(-1)^nS_n(n)$$
which completes the question with $$S_n(n)=n! (-1)^n$$
