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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
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up vote 1 down vote accepted

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

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

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