Sum over k of m choose k by -1^m-k by k to the n

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Theorem

Let $m, n \in \Z_{\ge 0}$.

$\displaystyle \sum_k \binom m k \left({-1}\right)^{m - k} k^n = m! \left\{ {n \atop m}\right\}$

where:

$\dbinom m k$ denotes a binomial coefficient
$\displaystyle \left\{ {n \atop m}\right\}$ etc. denotes a Stirling number of the second kind
$m!$ denotes a factorial.


Proof

The proof proceeds by induction on $m$.


For all $m \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:

$\displaystyle \forall n \in \Z_{\ge 0}: \sum_k \binom m k \left({-1}\right)^{m - k} k^n = m! \left\{ {n \atop m}\right\}$


Basis for the Induction

$P \left({0}\right)$ is the case:

\(\displaystyle \) \(\) \(\displaystyle \sum_k \binom 0 k \left({-1}\right)^{0 - k} k^n\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_k \delta_{0 k} \left({-1}\right)^{- k} k^n\) Zero Choose n
\(\displaystyle \) \(=\) \(\displaystyle \left({-1}\right)^{- 0} 0^n\) all terms vanish except for $k = 0$
\(\displaystyle \) \(=\) \(\displaystyle \delta_{0 n}\) $0^n = 0$ except when $n = 0$, as $0^0 = 1$
\(\displaystyle \) \(=\) \(\displaystyle 0! \left\{ {n \atop 0}\right\}\) Definition of Stirling Numbers of the Second Kind

So $P \left({0}\right)$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that, if $P \left({r}\right)$ is true, where $r \ge 1$, then it logically follows that $P \left({r + 1}\right)$ is true.


So this is the induction hypothesis:

$\displaystyle \sum_k \binom r k \left({-1}\right)^{r - k} k^n = r! \left\{ {n \atop r}\right\}$


from which it is to be shown that:

$\displaystyle \sum_k \binom {r + 1} k \left({-1}\right)^{r + 1 - k} k^n = \left({r + 1}\right)! \left\{ {n \atop r + 1}\right\}$


Induction Step

This is the induction step:

\(\displaystyle \) \(\) \(\displaystyle \sum_k \binom {r + 1} k \left({-1}\right)^{r + 1 - k} k^n\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_k \left({\binom r k + \binom r {k - 1} }\right) \left({-1}\right)^{r + 1 - k} k^n\) Pascal's Rule
\(\displaystyle \) \(=\) \(\displaystyle \sum_k \binom r k \left({-1}\right)^{r + 1 - k} k^n + \sum_k \binom r {k - 1} \left({-1}\right)^{r + 1 - k} k^n\)
\(\displaystyle \) \(=\) \(\displaystyle -\sum_k \binom r k \left({-1}\right)^{r - k} k^n + \sum_k \binom r {k - 1} \left({-1}\right)^{r - \left({k - 1}\right)} k^n\)
\(\displaystyle \) \(=\) \(\displaystyle -r! \left\{ {n \atop r}\right\} + \sum_k \binom r {k - 1} \left({-1}\right)^{r - \left({k - 1}\right)} k^n\) Induction Hypothesis
\(\displaystyle \) \(=\) \(\displaystyle -r! \left\{ {n \atop r}\right\} + \sum_k \binom r k \left({-1}\right)^{r - k} \left({k + 1}\right)^n\) Translation of Index Variable of Summation



So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_k \binom m k \left({-1}\right)^{m - k} k^n = m! \left\{ {n \atop m}\right\}$

$\blacksquare$


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