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I wonder how to elegantly prove the following identity for all $n\ge 3$:

$$\frac{n}{2}\cdot \sum_{x+y=n-1,x\ge 1,y\ge 1}\binom{n-1}{x}x^{x-1}y^{y-1} = n\cdot(n-2)\cdot (n-1)^{n-3} $$

via the binomial theorem. (A combinatorial proof is not too difficult; both quantities count the number of rooted labeled trees with root degree $2$, see here: rooted labeled trees with root degree 2).

Binomial theorem should be easy as well, since the LHS is actually almost in the correct form.

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    The LHS is most definitely not in the right form to apply binomial theorem onto it. The bases of the numbers with powers must be constants with respect to the sum. – Simply Beautiful Art Jul 27 '16 at 13:01
  • That's a quite good point. – mathse Jul 27 '16 at 13:37
  • Multiplying by $2/n$ we see that the claim is equivalent to saying that $$[z^{n-1}] T(z)^2 = 2(n-2) (n-1)^{n-3}$$ where $T(z)$ is the labeled tree function. This has appeared several times at MSE and a proof can be found at this MSE link. It also follows combinatorially from the species $$\mathcal{Z}\mathcal{T}^2.$$ – Marko Riedel Jul 27 '16 at 20:56

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