I hope this is a suitable MO question. In a research project, my collaborator and I came across some combinatorial expressions. I used my computer to test a few numbers and the pattern was suggesting the following equation for fixed integers $K\geq n>0$.


$$\dfrac{K!}{n!K^{K-n}}\sum\limits\_{ \begin{subarray}{c} k\_1+\dotsb+k\_{n}=K \\ k\_i \geq 1 \end{subarray}}
\prod\limits\_{i=1}^n \dfrac{k\_i^{k\_i-2}}{(k\_i-1)!}=\displaystyle {K-1\choose n-1}.$$


We tried to think of a proof but failed. One can probably move these $K!, n!$ to the right and rewrite the RHS, or maybe move $K!$ into the summation to form combinatorial numbers like $K\choose k\_1,k\_2,\dotsc,k\_n$. We don't know which is better. 


The questions are:


1. Anyone knows a proof for this identity?
2. In fact the expression that appears in our work is $\sum\limits\_{ \begin{subarray}{c} k\_1+\dotsb+k\_{n}=K \\ k\_i \geq 1 \end{subarray}}
\sigma\_p(k\_1,\dotsc,k\_n) \prod\limits\_{i=1}^n \dfrac{k\_i^{k\_i-2}}{(k\_i-1)!}$, where $p$ is a fixed integer and $\sigma\_p(\dotsc)$ is the $p$-th elementary symmetric polynomial. The equation in the beginning simplifies this expression for $p=0,1$. Is there a similar identity for general $p$?


----------Update----------


Question 2 is perhaps too vague, and I'd like to make it a bit more specific. Probably I should have written this down in the beginning, but I feared this is too long and unmotivated. But after seeing people's skills, I'm very tempted to leave it here in case somebody has remarks. In fact, question 2 partly comes from the effort to find a proof for the following (verified by computer).


$$
\frac{1}{K!} \prod\_{r=1}^{K} (r+1 -x)=
\sum\_{n=1}^K \frac{(-1)^n}{n!} \left[ \sum\_{p=0}^n K^{n-p} \prod\_{r=1}^p (x +r-4) 
 \left( 
\sum\limits\_{ \begin{subarray}{c} k\_1+\dotsb+k\_{n}=K \\ k\_i \geq 1 \end{subarray}}
\sigma\_p(k\_1,\dotsc,k\_{n}) \prod\limits\_{i=1}^n \dfrac{k\_i^{k\_i-2}}{(k\_i-1)!}
 \right) \right],
$$
Where $x$ is a fixed number (in our case, an integer).



A combinatorial identity
This is the answer to the first question, I wrote a long answer to Question 2 as a separate answer.


Note that $A:=\sum\_{k\_i>0,k\_1+\dots+k\_n=K}\frac{K!}{n!k\_1!\dots k\_n!} \prod k\_i^{k\_i-1}$ is a number of forests on the ground set $\{1,2,\dots,K\}$ having exactly $n$ connected components and with a marked vertex in each component ($k\_i$ correspond to the sizes of components.) Add a vertex 0 and join it with the marked vertices. Then we have to count the number of trees on $\{0,1,\dots,K\}$ in which $0$ has degree $n$. Remember that the sum of $z\_0^{d\_1-1}\dots z\_K^{d\_K-1}$ over all trees on $\{0,\dots,K\}$, where $d\_i$ is degree of $i$, equals $(z\_0+\dots+z\_K)^{K-1}$. Substitute $z\_{1}=\dots=z\_K=1$ and take a coefficient of $z\_0^{n-1}$. It is $\binom{K-1}{n-1}K^{K-n}$.


