Showing Discrete Sum Equality

Question

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.

Answer 1

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?

Answer 2

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.