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 $k(k-1){n\choose k}$ ?
 A: 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
$(7)$: $\binom{2n-3}{n-3}\stackrel{(1)}=\binom{2n-3}{n}\stackrel{(2)}=\frac{2n-3}{n}\binom{2n-4}{n-1}\stackrel{(1)}=\frac{2n-3}{n}\binom{2n-4}{n-3}\stackrel{(2)}=\frac{n-2}{n}\binom{2n-3}{n-2}$
$(7)$: $\binom{2n-2}{n-2}\stackrel{(1)}=\binom{2n-2}{n}\stackrel{(2)}=\frac{2n-2}{n}\binom{2n-3}{n-1}\stackrel{(1)}=\frac{2n-2}{n}\binom{2n-3}{n-2}$  
A: There's a factor of $n-1$ missing from the right side of the equation. Let us write the equation as 
$$\sum_{k=1}^n k (k-1){\binom{n}{k}}\cdot k{\binom{n}{n-k}}=n^2(n-1)\binom{2n-3}{n-2}.$$
Now count the number of sequences of length $2n$ on the alphabet $\{a_0,a_1,a_2,b_0,b_1\}$ with $n$ $a$'s (with any subscript) and $n$ $b$'s (with any subscript)  satisfying the following additional conditions. There is exactly one $a_1$, one $a_2$ and one $b_1$. Furthermore, $a_1$ and $a_2$ appear in the first half of the sequence and $b_1$ appears in the second half.
A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{\sum_{k = 1}^{n}k^{2}\pars{k - 1}{n \choose k}^{2} =
     n^{2}\pars{n - 1}{2n - 3 \choose n - 2}}$

\begin{align}
&\mbox{Lets consider}\quad
\fermi\pars{x} \equiv \sum_{k = 0}^{n}{n \choose k}^{2}x^{k}
\\&\mbox{such that}\quad
\sum_{k = 1}^{n}k^{2}\pars{k - 1}{n \choose k}^{2}
=\left.\bracks{\pars{x\,\partiald{}{x}}^{3} - \pars{x\,\partiald{}{x}}^{2}}\fermi\pars{x}
\right\vert_{x\ =\ 1}\tag{1}
\end{align}

Hereafter we'll use the identity
$$\color{#c00000}{%
{m \choose n} = \int_{\verts{z} = 1}{\pars{1 + z}^{m} \over z^{n + 1}}
\,{\dd z \over 2\pi\ic}\,,\qquad m, n \in {\mathbb N}\,,\quad m \geq n}\tag{2}
$$

\begin{align}
\fermi\pars{x}&=\sum_{k = 0}^{n}x^{k}{n \choose k}
\int_{\verts{z} = 1}{\pars{1 + z}^{n} \over z^{k + 1}}\,{\dd z \over 2\pi\ic}
=\int_{\verts{z} = 1}{\pars{1 + z}^{n} \over z}
\sum_{k = 0}^{n}{n \choose k}\pars{x \over z}^{k}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\int_{\verts{z} = 1}{\pars{1 + z}^{n} \over z}\pars{1 + {x \over z}}^{n}
\,{\dd z \over 2\pi\ic}
=\int_{\verts{z} = 1}{\pars{1 + z}^{n} \over z^{n + 1}}\,\pars{x + z}^{n}
\,{\dd z \over 2\pi\ic}\tag{3}
\end{align}

With $\pars{1}$ and $\pars{3}$ we'll have:
\begin{align}
&\color{#00f}{\large\sum_{k = 1}^{n}k^{2}\pars{k - 1}{n \choose k}^{2}}
\\[3mm]&=\int_{\verts{z} = 1}{\pars{1 + z}^{n} \over z^{n + 1}}\,
n\pars{n - 1}\bracks{%
\pars{n - 2}\pars{1 + z}^{n - 3} + 2\pars{1 + z}^{n - 2}}\,{\dd z \over 2\pi\ic}
\\[3mm]&=n\pars{n -1}\pars{n - 2}
\int_{\verts{z} = 1}{\pars{1 + z}^{2n - 3} \over z^{n + 1}}\,{\dd z \over 2\pi\ic}
+
2n\pars{n -1}
\int_{\verts{z} = 1}{\pars{1 + z}^{2n - 2} \over z^{n + 1}}\,{\dd z \over 2\pi\ic}
\\[3mm]&=n\pars{n -1}\pars{n - 2}{2n - 3 \choose n}
+
2n\pars{n -1}{2n - 2 \choose n}
\\[3mm]&={\pars{2n - 3}! \over \pars{n - 3}!\pars{n - 3}!}
+ 2\,{\pars{2n - 2}! \over \pars{n - 2}!\pars{n - 2}!}
\\[3mm]&={\pars{n - 1}\pars{n - 2}^{2} + 2\pars{n - 1}\pars{2n - 2}
\over \pars{n - 2}!\pars{n - 1}!}\,\pars{2n - 3}!
=\bracks{\pars{n - 2}^{2} + 4n - 4}\pars{n - 1}
{2n - 3 \choose n - 2}
\\[3mm]&=\color{#00f}{\large n^{2}\pars{n - 1}{2n - 3 \choose n - 2}}
\end{align}
