We can prove this using the labelled tree
function that is known from combinatorics.
Suppose we seek to verify that
$$2(n-1) n^{n-2} =
\sum_{k=1}^{n-1} {n\choose k} k(n-k) Q_k Q_{n-k}$$
where $Q_k$ is the number of unrooted labeled trees on $k$ nodes.
This is the same as evaluating
$$\sum_{k=1}^{n-1} {n\choose k} T_k T_{n-k}$$
where $T_k$ is the number of rooted labeled trees on $k$ nodes.
We have multiplied $Q_k$ by $k$, reflecting the $k$ choices for
the root.
We will provide a closed form of the exponential generating function
of the product of the two terms that are involved.
The combinatorial class of rooted labelled trees has the specification
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}\mathcal{T} =
\mathcal{Z} \times \textsc{SET}(\mathcal{T})$$
which gives the functional equation
$$T(z) = z \exp T(z).$$
Observe that when we multiply two exponential generating functions of
the sequences $\{a_n\}$ and $\{b_n\}$ we get that
$$ A(z) B(z) = \sum_{n\ge 0} a_n \frac{z^n}{n!}
\sum_{n\ge 0} b_n \frac{z^n}{n!}
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{1}{k!}\frac{1}{(n-k)!} a_k b_{n-k} z^n\\
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{n!}{k!(n-k)!} a_k b_{n-k} \frac{z^n}{n!}
= \sum_{n\ge 0}
\left(\sum_{k=0}^n {n\choose k} a_k b_{n-k}\right)\frac{z^n}{n!}$$
i.e. the product of the two generating functions is the generating
function of $$\sum_{k=0}^n {n\choose k} a_k b_{n-k}.$$
Now since $[z^0] T(z) = 0$ we may write for the convolution in
question that it is
$$\sum_{k=0}^{n} {n\choose k} T_k T_{n-k}.$$
Therefore we are dealing with the scenario
$$A(z) = B(z) = T(z).$$
The equality that we seek to prove is the convolution of the two
exponential generating functions $A(z)$ and $B(z)$ and to verify it we
must show that
$$n! [z^n] A(z) B(z) = 2 (n-1) n^{n-2}$$
We thus compute
$$n! [z^n] A(z) B(z)
= n! \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} T(z)^2 dz.$$
Put $T(z) = w$ to get $z=w\exp(-w)$ and $dz = (\exp(-w)-w\exp(-w)) \; dw$
to obtain
$$n! \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{\exp(w(n+1))}{w^{n+1}}
\times w^2\times (\exp(-w) - w\exp(-w)) \; dw
\\ = n! \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{\exp(wn)}{w^{n-1}} (1-w) \; dw
\\ = n!
\left(\frac{n^{n-2}}{(n-2)!} - \frac{n^{n-3}}{(n-3)!}\right).$$
This is
$$n^{n-2} (n(n-1)-(n-1)(n-2))
= n^{n-2} (n^2-n-n^2+3n-2)
\\ = n^{n-2} \times (2n-2)$$
as claimed.
The labelled tree function recently appeared at this
MSE link.
