# Prove ∑pq=α(qα)(pq)(−1)q(−q)pqα=p!α!.\sum_{q=\alpha}^p \binom{q}{\alpha} \binom{p}{q}\frac{(-1)^q(-q)^p}{q^\alpha}=\frac{p!}{\alpha!}.

How to prove

$\displaystyle \sum_{q=\alpha}^p \binom{q}{\alpha} \binom{p}{q}\frac{(-1)^q(-q)^p}{q^\alpha}=\frac{p!}{\alpha!}$

for $1 \leq \alpha \leq p$?

EDIT: This is a result that I derived after playing around with the (given) fact that

$\displaystyle\sum_{q=1}^p\sum_{j=1}^q q^{p-2}(1+h/q)^{j-1}\prod_{i=1,i\neq q}^p\frac{1}{q-i}=\sum_{q=1}^p[(1+h/q)^q-1]\frac{(-1)^{p-q}q^{p}}{q!(p-q)!}=\sum_{q=1}^p \frac{h^q}{q!}$

and grouping the coefficients of each $h^\alpha$ in both sides.

I have tried some values of $\alpha$ and $p$ and it works, so the formula seems to be true. I just don’t know how to proceed to prove this result.

Suppose we seek to verify that

We use the integral representation

which is zero when $q\lt r$ (pole vanishes) so we may extend $q$
back to zero.

We also use the integral

We thus obtain for the sum

We extract the residue on the inner integral to obtain

It remains to compute

Observe that $\exp(w)-1$ starts at $w$ so $(\exp(w)-1)^{p- r}$
starts at $w^{p- r}$ and hence only the constant coefficient from
$\exp( r w)$ contributes, the value being one, which finally
yields