I need to prove the following but I'm struggling to do so:
$$\left(r+1\right)\sum _{j=0}^n\:\frac{\left(j+r\right)!}{j!}=\frac{\left(n+r+1\right)!}{n!\:}$$
So firstly I considered the case when $n = 0$ and I got that $(r+1)!$ was on both the LHS and RHS so it was true for $n = 0$.
I assumed it was true when $n=k$ but I'm not sure what to do for the $n=k+1$ step
The left-hand side with $n=k+1$ is $$ (r+1)\sum_{j=0}^{k+1}\frac{(j+r)!}{j!}= (r+1)\sum_{j=0}^k\frac{(j+r)!}{j!}+(r+1)\frac{(k+1+r)!}{(k+1)!} $$ By the induction hypothesis, you can go on with $$ =\frac{(k+r+1)!}{k!}+(r+1)\frac{(k+1+r)!}{(k+1)!} $$ Collect the common terms and finish up.