Seemingly simple high school combinatorics proof doesn't add up Prove that $\binom{n}{n-2}\binom{n+2}{n-1}$ is an integer for all $n\in\mathbb Z^+$.
My take on this:
Recall: $${n\choose k} = \frac{n!}{k!(n-k)!}$$
So our problem reads (simplified): $$\frac{n!}{(n-2)!2!} \cdot \frac{(n+2)!}{(n-1)3!}$$
But simply substituting $n=1$ does not work for the problem (answer becomes $-\frac{1}{2}$) and therefore the original assumption must be wrong? Or am I misunderstanding?
I'm stuck in the proof:
Trying to multiply these fractions together I can get:
$$\frac{n!(n+2)!}{12(n-2)!(n-1)!}$$ and I can divide $n!$ by $(n-1)!$ to get $$\frac{n(n+2)!}{12(n-2)!}$$ but I'm stuck here. This still does not clearly show that the answer is an integer.

By definition $\dbinom{n}k=0$ when $k<0$, so 
$$\binom{1}{-1}\binom{3}0=0\cdot1=0\;.$$
Added: In the edited question you’ve reduced the case $n\ge 2$ to showing that
$$\frac{n(n+2)!}{12(n-2)!}$$
is an integer. Do a bit more cancellation to get
$$\frac{n^2(n+2)(n+1)(n-1)}{12}\;.$$
Now show that any product of four consecutive integers is a multiple of $12$.
