I'm going to university in October and thought I'd have a go at a few questions from one of their past papers. I have completed the majority of this question but I'm stuck on the very last part. In honesty I've been working on this paper a while now and I'm a bit tired so I'm probably giving up earlier than I usually would.
I won't write out the full question, only the last part:
Let $$S_r(n) = \sum_{m=0}^n\ (-1)^m m^r {n \choose m}$$ where r is a non-negative integer . Show that $S_r(n)=0$ for $r<n$. Evaluate $S_n(n)$.
I have shown that $S_r(n)=0$ for $r<n$ by taking $(1+z)^n= \sum_{m=0}^n\ z^m {n \choose m}$, letting $D_r(f(z))=z\frac d{dz}(z\frac d{dz}...(\frac d{dz}(f(z)))...)$ where $z\frac d{dz}$ is applied $r$ times and applied it to both sides. The left hand side gives a polynomial, degree n which has factor $(1+z)$ for all $r<n$ and the right hand side gives $\sum_{m=0}^n\ z^m (m)^r {n \choose m}$. Setting $z=-1$ yields the required result.
There is some build up to this, so I'm fairly certain that this was the intended method.
I'm stuck however on the very last part. I have tried finding a form for $D_r((1+z)^n)$, but I'm fairly sure that this isn't the correct approach, as the wording of the question implies that $S_n(n)$ needs to be considered separately.
I'm surprised that I didn't find that this question had already been asked, so apologies if it has been.
Thank you.