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.