Help with this combinatorial proof $\sum\limits_{k=1}^nk^2(k-1){n\choose k}^2 = n^2(n-1){2n-3\choose n-2}$ considering $n\ge2$ 
$\displaystyle\sum\limits_{k=1}^nk^2(k-1){n\choose k}^2 = n^2(n-1)
{2n-3\choose n-2}$ considering $n\ge2$

Can somebody help with this combinatorial proof?
I'm struggling a lot.
Thanks.
EDIT: Ok. I could figure it out, if we had $\displaystyle\sum\limits_{k=1}^nk^2{n\choose k}^2 = n^2
{2n-2\choose n-1}$.
The problem is, i don't understand what to do with that $(k-1)$ and how it leads to ${2n-3\choose n-2}$.
I know $k{n\choose k} = n{n-1\choose k-1}$ 
Choosing a team of $k$ elements from $n$ and from that $k$ elements, pick a captain is the same as choose a captain first, and then, complete the team, choosing $k-1$ elements from $n-1$
But, what about $k(k-1){n\choose k}$ ?

Hint: Note that because choosing $k$ elements from a set of $n$ is the same as choosing the complement of the $k$ elements, we have
$$
\binom{n}{k}=\binom{n}{n-k}\tag{1}
$$
and since choosing a team of $k$ people and then a leader from those chosen is the same as choosing a leader and then choosing the remaining $k-1$ from the remaining $n-1$, we get
$$
k\binom{n}{k}=n\binom{n-1}{k-1}\tag{2}
$$
and
$$
k^2(k-1)=k(k-1)(k-2)+2k(k-1)\tag{3}
$$
Then consider Vandermonde's Identity.

Full Solution:
$$
\hspace{-5mm}\begin{align}
&\sum_{k=1}^nk^2(k-1)\binom{n}{k}^2\\
&=\sum_{k=1}^nk(k-1)(k-2)\binom{n}{k}\binom{n}{n-k}+2\sum_{k=1}^nk(k-1)\binom{n}{k}\binom{n}{n-k}\tag{4}\\
&=n(n-1)(n-2)\sum_{k=1}^n\binom{n-3}{k-3}\binom{n}{n-k}+2n(n-1)\sum_{k=1}^n\binom{n-2}{k-2}\binom{n}{n-k}\tag{5}\\
&=n(n-1)(n-2)\binom{2n-3}{n-3}+2n(n-1)\binom{2n-2}{n-2}\tag{6}\\[4pt]
&=(n-1)(n-2)^2\binom{2n-3}{n-2}+4(n-1)^2\binom{2n-2}{n}\tag{7}\\[4pt]
&=n^2(n-1)\binom{2n-3}{n-2}\tag{8}
\end{align}
$$
Explanation:
$(4)$: apply $(1)$ and $(3)$
$(5)$: apply $(2)$ several times
$(6)$: Vandermonde Identity
$(7)$: $\binom{2n-3}{n-3}\stackrel{(1)}=\binom{2n-3}{n}\stackrel{(2)}=\frac{2n-3}{n}\binom{2n-4}{n-1}\stackrel{(1)}=\frac{2n-3}{n}\binom{2n-4}{n-3}\stackrel{(2)}=\frac{n-2}{n}\binom{2n-3}{n-2}$
$(7)$: $\binom{2n-2}{n-2}\stackrel{(1)}=\binom{2n-2}{n}\stackrel{(2)}=\frac{2n-2}{n}\binom{2n-3}{n-1}\stackrel{(1)}=\frac{2n-2}{n}\binom{2n-3}{n-2}$  
