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


for 1αp?

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


and grouping the coefficients of each hα in both sides.

I have tried some values of α 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<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

\frac{(p- r)!}{2\pi i}
\frac{1}{w^{p- r+1}}
{p\choose p- r} \exp( r w)
(\exp(w)-1)^{p- r}
\; dw
\\ = \frac{p!}{ r!}
\frac{1}{2\pi i}
\frac{1}{w^{p- r+1}}
\exp( r w)
(\exp(w)-1)^{p- r}
\; dw.

It remains to compute
[w^{p- r}]\exp( r w)
(\exp(w)-1)^{p- r}.

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

\frac{p!}{ r!}.

Source : Link , Question Author : Jeze Ken , Answer Author : Marko Riedel

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