On combinatoric equality $\sum_{i=1}^{n-1}{n\choose i} i^{i-1} (n-i)^{n-i-1}=2(n-1)n^{n-2}$

Question

This originates from Cayley's formula of tree/forest counting. I'm using notations from reference: https://www.cambridge.org/core/services/aop-cambridge-core/content/view/94150E13E32551CF87B0326FA458AEA1/S144678870001274Xa.pdf/random-walks-on-random-trees.pdf.

The lemma 2.1 give a formula for $C(n,k)$, which defined as the number of trees with $n$ labelled nodes in which a given node has degree $k$. Consider the case $k=2$, the formula give $(n-2)(n-1)^{n-3}$. Another way to count the tree is to consider two subtrees after deleting the given node. For each bipartition of the rest $n-1$ nodes into $i$ and $n-i-1$ nodes. There is $i*i^{i-2}*(n-i-1)*(n-i-1)^{n-i-3}$ possible subtrees and assigning rooting, and there are $n-1\choose i$ ways of partition. So we would expect $2C(n,2)=\sum_{i=1}^{n-2}{n-1\choose i} i^{i-1} (n-i-1)^{n-i-2}$, replace $n$ with $n+1$ gives the fomular in title.

My question is: is it possible to prove this formula directly without introducing tree-counting? I've attempted considering generating function of $n^{n-1}$: $\sum_{i=1}^{\infty}\frac{n^{n-1}}{n!}x^n=-W(-x)$, in which $W$ is the Lambert W-Function. But not able to get any further. Are there any clue to do this?

Answer 1

A proof using the Lambert $W$ function could probably follow these lines.

We have $$ \sum_{i=1}^{n-1}\binom ni i^{i-1}(n-i)^{n-i-1}=[x^n] \left(\sum_{i\geq 1} \frac{i ^{i-1}}{i!}x^i\right)^2=[x^n]W(-x)^2. $$

But now integer powers of $W$ also have explicit Taylor expansions (see here: https://en.wikipedia.org/wiki/Lambert_W_function#Integer_and_complex_powers) which can be proved by Lagrange inversion formula. The details are hopefully easy to fill in.