# Trying to prove a binomial equation

I've been emptying my notebook over this, and still reach the same nothing at the end.
I'm trying to prove that the following equation is true, with no luck:

$\forall n,k \in \mathbb{N}^+ . k<n\to \binom{n-1}{k-1}\binom{n}{k+1}\binom{n+1}{k}=\binom{n-1}{k}\binom{n}{k-1}\binom{n+1}{k+1}$

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For a combinatorial proof, see this answer. – Mike Spivey Jan 21 '13 at 16:57

$$\binom{n-1}{k-1}\binom{n}{k+1}\binom{n+1}{k}$$

$$=\frac{(n-1)!}{(n-k)!(k-1)!}\frac{n!}{(k+1)!(n-k-1)!}\frac{(n+1)!}{k!(n+1-k)!}$$

$$=\frac{(n-1)!}{k!(n-1-k)!}\frac{n!}{(k-1)!\{n-(k-1)\}!}\frac{(n+1)!}{(k+1)!\{(n+1)-(k+1)\}!}$$ as $n-k-1=n-1-k,n+1-k=n-(k-1),n-k=(n+1)-(k+1)$

$$=\binom{n-1}k\binom n{k-1}\binom{n+1}{k+1}$$

We know, $\binom m r=0$ if $r<0$ or if $r>m\implies \binom m r>0$ if $0<r<m$

So,

$\binom{n-1}k>0$ if $0\le k\le n-1$

$\binom n{k-1}>0$ if $0\le k-1\le n\iff 1\le k\le n+1$

$\binom {n+1}{k+1}>0$ if $0\le k+1\le n+1\iff 0\le k\le n$

All the three will be $>0$

if $$\text{max}(0,1,0)\le k\le \text{min}(n-1,n+1,n)\implies 1\le k\le n-1$$

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