<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] The product is:
$$\prod (n^2-k^2)=(n-r)(n-r+1)\cdots(n-1)(n+1)\cdots(n+r-1)(n+r)$$
$$=\frac{(2r+1)!}{n}{n+r \choose 2r+1}$$
The $n$ is because we are missing that term, the binomial symbol is because we recognize a product of $2r+1$ going down from $n+r$, and the factorial is because the binomial symbol is missing its denominator.
You get
$$\sum_{r=0}^\infty (-1)^r\frac{(2r+1)!}{r!(r+1)!}{n+r \choose 2r+1}$$
Here you recognize another choose symbol
$$\sum_{r=0}^\infty (-1)^r{2r+1\choose r}{n+r \choose 2r+1}=\sum_{r=0}^{n-1} (-1)^r{2r+1\choose r}{n+r \choose 2r+1}$$
where the last term told us the largest reasonable value for $r$ (if bottom is larger than top, it's zero).
You already got to here. Now you can just calculate the recursion. Assuming you have term $a_n$, let's compute the next one:
$$a_{n+1}=\sum_{r=0}^{n} (-1)^r{2r+1\choose r}{n+1+r \choose 2r+1}$$
We use the summation property of the Pascal's triangle:
$$a_{n+1}=\sum_{r=0}^{n} (-1)^r{2r+1\choose r}\left({n+r \choose 2r+1}+{n+r \choose 2r}\right)$$
Can you continue expressing this with $a_n$?
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