infinite series

General term of this sum =

\frac{n^2(n^2-1^2)(n^2-2^2)\cdots (n^2-r^2)}{(r+1)!}
=\frac{(n+r)(n+r-1)\cdots (n+1)(n)(n)(n-1)\cdots (n-r+1)(n-r)}{(r+1)!^2}
=\binom{n+r}{r+1}\binom{n}{r+1}

(1-x^{-1})^{-n}=1+nx^{-1}+\binom{n+1}{2}x^{-2}+\cdots

(1-x)^{n}=1-nx+\binom{n}{2}x^2-\cdots

Therefore the constant term in
(1-x^{-1})^{-n}\cdot (1-x)^{n} \text{ is } =1-n^2+\frac{n^2(n^2-1^2)}{2}-\cdots

\text{But }(1-x^{-1})^{-n}\cdot (1-x)^{n}=(-x)^n\text{ has no constant term.}

\therefore 1-n^2+\frac{n^2(n^2-1^2)}{2}-\cdots =0