Showing Discrete Sum Equality The questions asks to show that 
$\sum\limits_{k=0}^n (-1)^{k}{n \choose k}(n-k)^{r} = \begin{cases}0 &   r=1,2,3,...,n-1 \\ n! &  r = n  \end{cases}$
Any help would be appreciated. I'm stumped on where to start. 
 A: If $f(x)=e^{(n-k)x}$, then $f'(x)=(n-k)e^{(n-k)x}$ and more generally $f^{(r)}(x)=(n-k)^re^{(n-k)x}$.
Thus
$$ \sum_{k=0}^n(-1)^k{n\choose k}(n-k)^r=\frac{\mathrm d^r}{\mathrm dx^r}\left.\sum_{k=0}^n(-1)^k{n\choose k}e^{(n-k)x}{}\right|_{x=0}$$
Now note that $e^{(n-k)x}=(e^x)^{n-k}$ so that the sum simplifies to $\left(e^x-1\right)^n$. Can you compute the first $n$ derivatives of this?
A: Observe that when we  multiply two exponential generating functions of
the sequences $\{a_n\}$ and $\{b_n\}$ we get that
$$ A(z) B(z) = \sum_{n\ge 0} a_n \frac{z^n}{n!} 
\sum_{n\ge 0} b_n \frac{z^n}{n!}
= \sum_{n\ge 0} 
\sum_{k=0}^n \frac{1}{k!}\frac{1}{(n-k)!} a_k b_{n-k} z^n\\
= \sum_{n\ge 0} 
\sum_{k=0}^n \frac{n!}{k!(n-k)!} a_k b_{n-k} \frac{z^n}{n!}
= \sum_{n\ge 0} 
\left(\sum_{k=0}^n {n\choose k} a_k b_{n-k}\right)\frac{z^n}{n!}$$
i.e. the  product of  the two generating  functions is  the generating
function of $$\sum_{k=0}^n {n\choose k} a_k b_{n-k}.$$
In the present case we have
$$A(z) = \exp(-z)$$ and
$$B(z) = \sum_{q\ge 0} q^r \frac{z^q}{q!}.$$
Using the Stirling numbers of the second kind $B(z)$ simplifies to
$$B(z) = \sum_{q\ge 0} 
\frac{z^q}{q!}
\sum_{p=0}^r {r\brace p} q^{\underline{p}}
= \sum_{p=0}^r {r\brace p}
\sum_{q\ge 0} 
q^{\underline{p}} \frac{z^q}{q!}
\\ = \sum_{p=0}^r {r\brace p}
\sum_{q\ge p} 
q^{\underline{p}} \frac{z^q}{q!}
= \sum_{p=0}^r {r\brace p}
z^p \sum_{q\ge p} 
q^{\underline{p}} \frac{z^{q-p}}{q!}
\\ = \sum_{p=0}^r {r\brace p}
z^p \sum_{q\ge p} \left(\frac{d}{dz}\right)^p 
\frac{z^q}{q!}.$$
We may extend the inner sum back to $q$ equal zero because these terms
drop out when the differentiation is applied, getting
$$B(z) = \sum_{p=0}^r {r\brace p}
z^p \sum_{q\ge 0} \left(\frac{d}{dz}\right)^p 
\frac{z^q}{q!}
= \sum_{p=0}^r {r\brace p}
z^p \left[ \left(\frac{d}{dz}\right)^p \exp(z) \right]
\\= \exp(z) \sum_{p=0}^r {r\brace p} z^p.$$
We thus finally obtain
$$n! [z^n] A(z) B(z)
= n! [z^n] \sum_{p=0}^r {r\brace p} z^p
= n! {r\brace n}.$$
This gives zero for $r\lt n$ and $n!$ for $r=n.$
Addendum. I just realized that among the links there is an answer of mine which is very similar to the above. I am posting anyway because the new version is quite a bit simpler than the induction proof in the old one.
