Let´s consider the complete graph $K_{n}$. Let $c_n(k)$ be equal to the number of spanning trees T with $deg_T(n)=k$.
First I want to prove the recursion $c_n(k)=\frac{k(n-1)}{n-1-k}c_n(k+1)$ and then from that I want to derive Cayley´s formula $n^{n-2}$.
Considering the first point I don´t get my head around finding a clear explanation to why this recursion holds.
Am I correct in assuming that the number of (labeled) spanning trees with $deg_T(n)=k$ are all the spanning trees for which the degree of the node n is equal to k?
I tried looking at the labeled spanning trees of $K_4$ and then I ordered them by the degree of the node 4, finding $c_4(1)=9, c_4(2)=6, and c_4(3)=1$. Adding those all together we get in fact all the labeled spanning trees of $K_4$, which is not surprising.
As mentioned I was not able to explain the stated recursion.
Concerning the second point I tried calculating $c_4(1)=\frac{1(n-1)}{n-1-1}c_4(2)$ and then evaluate $c_4(2)=\frac{2(n-1)}{n-1-2}c_4(3)$.
We know that $c_4(3)=1$ and now we can introduce our $c_4(2)=\frac{2(n-1)}{n-3}$ back into $c_4(1)=\frac{n-1}{n-1-1}c_4(2)$ so we get $c_4(1)=\frac{n-1}{n-2}\frac{2(n-1)}{n-3}$.
At this point I am stuck since I dont know how to get to $n^{n-2}$. I know that our recursion gives us all labeled spanning trees for $K_4$ but I can´t derive Cayley´s formula from the recursion.
I was searching for proofs of Cayleys formula on the internet and in books such as Proofs from The Book by Aigner and Ziegler but nowhere I found one that used the degree of the heaviest node.