$$\sum_{m=0}^q(n-m)\frac{(p+m)!}{m!}=\frac{(p+q+1)!}{q!}\left(\frac{n}{p+1}-\frac{q}{p+2}\right)$$
I thought of solving it by induction but am unable to fathom which variable should I apply induction principle upon. I thought of applying on q, but I am stuck. Please help.
We'll proceed by induction on $q$. The statement can be verified for $q=1$ (though as the comments show, it takes a little effort). Suppose it is known up to $q-1$, let's address it for $q$.
The left hand is, by induction, $$\frac {(p+q)!}{(q-1)!}\times \left(\frac n{p+1}-\frac {q-1}{p+2}\right)+(n-q)\frac {(p+q)!}{q!}$$
Grouping the terms involving $n$ we get $$n\times \frac {(p+q+1)!}{q!(p+1)}$$
Grouping the terms without $n$ we get $$-\frac {(p+q+1)!}{(q-1)!}\times \frac 1{p+2}$$
Adding these together quickly reduces to the right hand.