<s>[INST] Prove that $\sum\limits_{r=0}^\infty\,\dfrac{(-1)^r\,n}{r!\,(r+1)!}\,\prod\limits_{k=1}^r\,\left(n^2-k^2\right)=(-1)^{n+1}$ 
If n is a positive integer, prove that:
  $$\begin{align}n - {\frac{n(n^2-1^2)}{2!}}+{\frac{n(n^2-1^2)(n^2-2^2)}{2!3!}} + \ldots\phantom{aaaaaaaaaa} &&\\
+(-1)^r{\frac{n(n^2-1^2)(n^2-2^2)\cdots(n^2-r^2)}{r!(r+1)!}}+\ldots &=& (-1)^{n+1}\end{align}$$ 

I tried by simplifying it to $\binom{n}{1}\binom{n}{0}-\binom{n}{2}\binom{n+1}{1}+\binom{n}3\binom{n+2}{2}-\ldots$  , or $\sum(-1)^r\binom{n}{r+1}\binom{n+r}{r}$. But I don't know what to do from here.
 [/INST] $$(-1)^r{\frac{n(n^2-1^2)(n^2-2^2)...(n^2-r^2)}{r!(r+1)!}}=\dfrac{n(n-1)(n-2)\cdots(n-r)}{(r+1)!}\cdot\dfrac{-(n+1)\cdot-(n+2)\cdots-(n+r)}{r!}$$
Think of the coefficient of $x$ in $$(1+x)^n\left(1+\dfrac1x\right)^{-(n+1)}=\dfrac{x^{n+1}}{1+x}$$
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