An identity on the number of trees Let $T_n$ be the number of labelled trees on $n$ vertices, then 
$$ T_n=\sum_kk\binom{n-2}{k-1}T_kT_{n-k} \tag{1}$$
Using this question, I was able to prove that 
$$ T_n= \frac{n}{2} \  \sum\binom{n-2}{k-1}T_kT_{n-k} .$$
But I don't know how to prove $(1)$. Can anyone help me please?
 A: I incidentally came on this post. The OP was on the right path. He proved that $$T_n=\frac{n}{2}\sum_k\binom{n-2}{k-1}T_kT_{n-k}.$$ This is euqivalent to say $$\begin{eqnarray}2T_n&=&\sum_k(k+(n-k))\binom{n-2}{k-1}T_kT_{n-k}\\ &=&\sum_k\left(k\binom{n-2}{k-1}T_kT_{n-k}+(n-k)\binom{n-2}{n-k-1}T_{n-k}{T_k}\right)\\&=&\sum_kk\binom{n-2}{k-1}T_kT_{n-k}+\sum_jk\binom{n-2}{j-1}T_jT_{n-j}\\ &=&2\sum_kk\binom{n-2}{k-1}T_nT_{n-k}.\end{eqnarray}$$ From which the result follows.
A: The equation  under consideration can  be proved using the  species of
labelled rooted trees and its  functional equation. (Note: we will use
the letter $T$ to refer  to rooted labelled trees and their generating
function and  $Q$ to unrooted ones  in order to  adhere to established
convention.)
Now the species $\mathcal{T}$ of rooted labelled trees is given by
$$\mathcal{T} = \mathcal{Z} \times \mathfrak{P}(\mathcal{T}).$$
This immediately gives the classic functional equation 
$$T(z) = z e^{T(z)}$$
where $$T(z) = \sum_{n\ge 1} T_n \frac{z^n}{n!}.$$
Differentiate to obtain a recurrence, getting
$$T'(z) = e^{T(z)} + z e^{T(z)} T'(z)
= \frac{T(z)}{z} + T(z) T'(z).$$
Now observe that
$$T'(z) = \sum_{n\ge 0} T_{n+1} \frac{z^n}{n!}
\quad\text{and}\quad
\frac{T(z)}{z} = \sum_{n\ge 0} \frac{T_{n+1}}{n+1} \frac{z^n}{n!}.$$
Comparing coefficients  in the differentiated  functional equation and
using the convolution of exponential generating functions we find that
$$T_{n+1} =  \frac{T_{n+1}}{n+1}
+ \sum_{k=0}^{n-1} {n\choose k} T_{k+1} T_{n-k}.$$
Now switch  to unrooted  trees noting that  for the count $Q_n$ of unrooted
labelled trees we have
$$n \times Q_n = T_n$$ to obtain that
$$(n+1) Q_{n+1} =  Q_{n+1}
+ \sum_{k=0}^{n-1} {n\choose k}\times (k+1) Q_{k+1}\times (n-k) Q_{n-k}.$$
This yields
$$Q_{n+1} = \frac{1}{n} 
\sum_{k=1}^n {n\choose k-1} \times k Q_k \times (n+1-k) Q_{n+1-k}
\\= \sum_{k=1}^n \frac{(n-1)!}{(k-1)!(n-k)!} \times k Q_k \times Q_{n+1-k}
\\ = \sum_{k=1}^n {n-1\choose k-1} \times k Q_k \times Q_{n+1-k}.$$
This is what we were trying to prove (replacing $n+1$ by $n$), QED.
