Summation of series involving factorials. I got this question in a maths contest archive and I am completely clueless over how to start.
$$\sum_{m=0}^q(n-m){(p+m)!\over m!}= {(p+q+1)! \over q!}\left(\frac{ n}{ p+1}-\frac {q}{p+2}\right)$$
I thought of transforming $\frac{(p+m)!}{m!}$ into $p+m\choose m$ by multiplying and dividing by $p!$ but that was surely a bad idea or maybe I couldn't figure it out right.

For $q=0$ you have only the term $m=0$,$n\frac{p!}{0!}=n p!$ on the left side of the equation. On the right side you have $\frac{(p+1)!}{0!}\frac{n}{p+1}$. Noting that $(p+1)!=p!(p+1)$ you get the left hand side term.
We now prove by induction. We assume that the given equation is valid for $q$, and we want to prove that is valid for $q+1$.
$$\sum_{m=0}^{q+1}(n-m)\frac{(p+m)!}{m!}=\sum_{m=0}^{q}(n-m)\frac{(p+m)!}{m!}+(n-q-1)\frac{(p+q+1)!}{(q+1)!} = \frac{(p+q+1)!}{q!}\left(\frac{n}{p+1}-\frac{q}{p+2}\right)+(n-q-1)\frac{(p+q+1)!}{(q+1)!}=\frac{(p+q+1)!}{(q+1)!}\left(\frac{n(q+1)}{p+1}-\frac{q(q+1)}{p+2}+(n-q-1)\right)=\frac{(p+q+1)!}{(q+1)!}\left(\frac{n(q+1)+n(p+1)}{p+1}-\frac{q(q+1)+(p+2)(q+1)}{p+2}\right)=\frac{(p+q+2)!}{(q+1)!}\left(\frac{n}{p+1}-\frac{q+1}{p+2}\right)$$
Which is exactly what we wanted to show
