An identity on binomial coefficients without using explicit formulas

I need to prove that

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

without using $\ {n \choose k} = \frac{n!}{k!(n-k)!}$

so I am supposed to either use the idea similar to proving that $ (1-1)^n = 2^n = \sum \ {n \choose k} $ or find something that both RHS and LHS counts.

Unfortunately, I can't show any attempt here because all I could think of is to apply recurrence relations several times to both sides hoping to be able to see what is going on.

Recall the absorption identity $$\binom{N}{K}=\frac{N}{K}\binom{N-1}{K-1}.$$

This can be proven by counting the number of committees consisting of $K$ members among $N$ candidates, of which one is president. Indeed, either you first choose your $K$ members, then your president, yielding $$\binom{N}{K}K$$ options, or you first choose your president, and the the remaining $N-1$ members, yielding $$N\binom{N-1}{K-1}$$ options. Equating both counts, then dividing by $K$ gives the required identity.

Thus, if $k\neq 0,1$, your identity is equivalent to $$\binom{n-1}{k-1}\frac{n}{k+1}\binom{n-1}{k}\frac{n+1}{k}\binom{n}{k-1}=\binom{n-1}{k}\frac{n+1}{k+1}\frac{n}{k}\binom{n-1}{k-1}\binom{n}{k-1},$$ which clearly holds. If $k=0,1$, the identity is obvious.