Prove that $\binom{n}{1}^2+2\binom{n}{2}^2+\cdots +n\binom{n}{n}^2=n\binom{2n-1}{n-1}$ 
Prove that
  $$
\binom{n}{1}^2+2\binom{n}{2}^2+\cdots + n\binom{n}{n}^2
= n \binom{2n-1}{n-1}.
$$

So
$$
\sum_{k=1}^n k \binom{n}{k}^2
= \sum_{k=1}^n k \binom{n}{k}\binom{n}{k}
= \sum_{k=1}^n n \binom{n-1}{k-1} \binom{n}{k}
= n \sum_{k=0}^{n-1} \frac{(n-1)!n!}{(n-k-1)!k!(n-k-1)!(k+1)!}
= n^2 \sum_{k=0}^{n-1} \frac{(n-1)!^2}{(n-k-1)!^2k!^2(k+1)}
=n^2 \sum_{k=0}^{n-1} \binom{n-1}{k}^2\frac{1}{k+1}.
$$
I do not know what to do with $\frac{1}{k+1}$, how to get rid of that.

$$\sum_{k=1}^{n}k\binom{n}{k}^{2}=n\sum_{k=1}^{n}\binom{n-1}{k-1}\binom{n}{n-k}=n\sum_{k=0}^{n-1}\binom{n-1}{k}\binom{n}{n-k-1}=n\binom{2n-1}{n-1}$$
Applying Vandermonde's identity in third equality.
