Seeking a combinatorial proof of $2n^{n-3} = \sum_{m=1}^{n-1}\binom{n-2}{m-1}m^{m-2}(n-m)^{n-m-2}$

Question

I need to prove the following using Combinatorial proof: (Not using math laws But finding two similar Combinatorial problems)

$$2n^{n-3} = \sum_{m=1}^{n-1}\binom{n-2}{m-1}m^{m-2}(n-m)^{n-m-2}$$

(original problem image)

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I was told that solving another problem may help, So I solved it and here it is.

The number of undirected trees with n vertices such that the edge 1-2 doesn't exist is: $(n-2)*n^{n-3}$.

Answer 1

As @MishaLavrov mentioned in the comments, both sides are different ways of counting the number of undirected trees with $n$ vertices containing the edge $1-2$.

We collect the results:

Number of undirected trees with $k$ vertices: $f(k) =: k^{k-2}$.

Number of undirected trees with $n$ vertices containing the edge $1-2$: $g(k) =: 2k^{k-3}$.

Now, on your right hand side, imagine breaking your tree of $n$ vertices into two trees: one with $m$ vertices, and one with $n-m$ vertices. Now apply $f(k)$ on those two trees, and sum over all possible values of $m$.