[tex]\displaystyle\sum_{r=1}^nr(r+1)\cdots(r+p-1)[/tex]
When [tex]n=1[/tex],
[tex]\displaystyle\sum_{r=1}^1r(r+1)\cdots(r+p-1)=1(1+1)(1+2)\cdots(1+p-2)(1+p-1)=p![/tex]
Meanwhile, you have on the right
[tex]\dfrac{(1)(1+1)(1+2)\cdots(1+p-2)(1+p-1)(1+p)}{p+1}=(1)(1+1)(1+2)\cdots(p-1)(p)=p![/tex]
so the equality holds for [tex]n=1[/tex].
Assume it holds for [tex]n=k[/tex], i.e. that
[tex]\displaystyle\sum_{r=1}^kr(r+1)\cdots(r+p-1)=\frac{k(k+1)(k+2)\cdots(k+p-1)(k+p)}{p+1}[/tex]
Now for [tex]n=k+1[/tex], you have
[tex]\displaystyle\sum_{r=1}^{k+1}r(r+1)\cdots(r+p-1)=\sum_{r=1}^kr(r+1)\cdots(r+p-1)+(k+1)(k+2)\cdots(k+1+p-1)[/tex]
[tex]=\displaystyle\frac{k(k+1)(k+2)\cdots(k+p-1)(k+p)}{p+1}+(k+1)(k+2)\cdots(k+1+p-2)(k+1+p-1)[/tex]
[tex]=\displaystyle\frac{k(k+1)(k+2)\cdots(k+p-1)(k+p)}{p+1}+(k+1)(k+2)\cdots(k+p-1)(k+p)[/tex]
[tex]=\left(\dfrac k{p+1}+1\right)(k+1)(k+2)\cdots(k+p-1)(k+p)[/tex]
[tex]=\dfrac{k+p+1}{p+1}(k+1)(k+2)\cdots(k+p-1)(k+p)[/tex]
[tex]=\dfrac{(k+1)(k+2)\cdots(k+p-1)(k+p)(k+p+1)}{p+1}[/tex]
as required.
