How to prove

p∑q=α(qα)(pq)(−1)q(−q)pqα=p!α!

for 1≤α≤p?

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

p∑q=1q∑j=1qp−2(1+h/q)j−1p∏i=1,i≠q1q−i=p∑q=1[(1+h/q)q−1](−1)p−qqpq!(p−q)!=p∑q=1hqq!

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.

**Answer**

Suppose we seek to verify that

(−1)pp∑q=r(pq)(qr)(−1)qqp−r=p!r!.

We use the integral representation

(qr)=(qq−r)=12πi∫|z|=ϵ(1+z)qzq−r+1dz

which is zero when q<r (pole vanishes) so we may extend q

back to zero.

We also use the integral

qp−r=(p−r)!2πi∫|w|=γexp(qw)wp−r+1dw.

We thus obtain for the sum

(−1)p(p−r)!2πi∫|w|=γ1wp−r+1×12πi∫|z|=ϵzr−1p∑q=0(pq)(−1)q(1+z)qzqexp(qw)dzdw=(−1)p(p−r)!2πi∫|w|=γ1wp−r+1×12πi∫|z|=ϵzr−1(1−1+zzexp(w))pdzdw=(−1)p(p−r)!2πi∫|w|=γ1wp−r+1×12πi∫|z|=ϵ1zp−r+1(−exp(w)+z(1−exp(w)))pdzdw=(p−r)!2πi∫|w|=γ1wp−r+1×12πi∫|z|=ϵ1zp−r+1(exp(w)+z(exp(w)−1))pdzdw.

We extract the residue on the inner integral to obtain

\frac{(p- r)!}{2\pi i}

\int_{|w|=\gamma}

\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}

\int_{|w|=\gamma}

\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

yields

\frac{p!}{ r!}.

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