Prove equality by Catalan number

Question

I try to prove equality $$\sum_{k=0}^{n} {n\choose k}^2 = {2n\choose n}$$ by using a relation of Catalan number $$C_n = \sum_{k=1}^{n}C_{k-1}C_{n-k}.$$ After substituting $$C_n = \frac{1}{n+1} {2n\choose n},$$ and some derivations, I arrive at $${2n\choose n} = \sum_{k=0}^{n} \frac{k(2k-2)!(2n-2k)!(n+1)}{(n-k+1)(n-k)!^2k!^2},$$ so I need to show $$\frac{k(2k-2)!(2n-2k)!(n+1)}{n-k+1} = n!^2.$$ Now I'm having trouble doing this step and I was wondering if there is something wrong with my previous steps.