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 $\binom{K-1}{n-1}K^{K-n}$.

Here's another proof. We first rewrite the identity (by setting $k_i=j_i+1$) as $$ \sum_{j_1+\cdots +j_n=K-n}\prod_{i=1}^n \frac{(j_i+1)^{j_i-1}}{j_i!} = n\frac{K^{K-n-1}}{(K-n)!}. \tag{1} $$

Let $F(x)$ be the formal power series satisfying $F(x)= e^{xF(x)}$. It is well known (and easily proved, e.g., by Lagrange inversion) that $$F(x)^n = \sum_{j=0}^\infty n(j+n)^{j-1}\frac{x^j}{j!}.\tag{2}$$ In particular, $$F(x) = \sum_{j=0}^\infty (j+1)^{j-1}\frac{x^j}{j!}.$$ So the left side of $(1)$ is equal to the coefficient of $x^{K-n}$ in $F(x)^{n}$, which by $(2)$ is equal to the right side of $(1)$.

Here is a generating-function proof of your conjectured identity (and an answer to question 2).

The main ingredient is a formula for the appearing symmetric sums.

Let $T(z)$ (the ``tree function'') be the formal power series satisfying $T(z)=z\,e^{T(z)}$.

If $F$ is a formal power series the coefficients of $G(z):=F(T(z))$ are given by (Lagrange inversion) $$[z^0]G(z)=[z^0] F(z) \mbox{ , } [z^k]G(z)=\tfrac{1}{k} [y^{k-1}] F^\prime(y)\,e^{ky} =[y^k](1-y)F(y)\,e^{ky}\mbox{ for } k\geq 1\;.$$ In particular (as is well known) $$T(z)=\sum_{n\geq 1}\frac{n^{n-1}}{n!}z^n \;\mbox{ and }\; \frac{T(z)}{1-T(z)}=\sum_{n\geq 1}\frac{n^{n}}{n!}z^n$$ Thus $T(z)\left(1+\frac{t}{1-T(z)}\right)=\sum_{n\geq 1} \frac{(1+tn)n^{n-1}}{n!}$. Therefore \begin{align*}S_{p,n}(K):&=\sum_{{k_1+\ldots +k_n=K \atop k_i\geq 1}} \sigma_p(k_1,\ldots,k_n)\prod_{i=1}^n \frac{k_i^{k_i-1}}{k_i!}\\ &=[t^p]\sum_{k_1+\ldots +k_n=K \atop k_i\geq 1} \prod_{i=1}^n \frac{(1+tk_i) k_i^{k_i-1}}{k_i!}\\ &=[t^p z^K]\, T(z)^n \left(1+\frac{t}{1-T(z)}\right)^n \end{align*} and $$S_{p,n}(K)={n \choose p} [z^K]\frac {T(z)^n}{(1-T(z))^p}={n \choose p}[y^K]\,y^n\, \frac{(1-y)}{(1-y)^p}\,e^{Ky}\;\;\;\;\;(*)$$

Now consider the sum ($m:=x-4$) $$R(K,m):=\sum_{n=0}^K\frac{(-1)^n}{n!}\bigg[\sum_{p=0}^n K^{n-p}\,\left(\prod_{r=1}^p (m+r)\right) S_{n,p}(K)\bigg]$$ Since $K\geq 1$ the sum remains unchanged if we start the summation at $n=0$ (all $S_{0,p}(K)$ are $0$). Using that and $(*)$ gives \begin{align*} R(K,m):&=\sum_{n=0}^K\frac{(-1)^n}{n!}\bigg[\sum_{p=0}^n K^{n-p}\,\left(\prod_{r=1}^p (m+r)\right) S_{n,p}(K)\bigg]\\ &=[y^K]\,(1-y) \sum_{n=0}^K\frac{(-1)^n}{n!}\sum_{p=0}^n K^{n-p}\,p!{m+p \choose p}{n \choose p}\frac{y^n}{(1-y)^p}\,e^{Ky}\\ &=[y^K]\,(1-y) \sum_{p=0}^K\sum_{n=p}^K\frac{(-1)^n}{(n-p)!} K^{n-p}{m+p \choose p}\frac{y^n}{(1-y)^p}\,e^{Ky}\\ &=[y^K]\,(1-y) \sum_{p=0}^K {m+p \choose p}\frac{(-1)^p y^p}{(1-y)^p}\bigg[\sum_{n=p}^K\frac{(-1)^{n-p}}{(n-p)!} K^{n-p}y^{n-p}\,e^{Ky}\bigg]\\ &=[y^K]\,(1-y) \sum_{p=0}^K {m+p \choose p}\frac{(-1)^p y^p}{(1-y)^p}\bigg[1 +\mathcal{O}(y^{K-p+1})\bigg]\\ \end{align*} where $\mathcal{O}(y^{K-p+1})$ denotes a formal power series which is a multiple of $y^{K-p+1}$.
Taking into account the respective factors $y^p$ the terms in these series do not contribute to the coefficient $[y^K]$. Therefore

\begin{align*} R(K,m)&=[y^K]\,(1-y) \sum_{p=0}^K {m+p \choose p}\frac{(-1)^p y^p}{(1-y)^p}\\ &=[y^K]\,(1-y) \sum_{p\geq 0} {m+p \choose p}\frac{(-1)^p y^p}{(1-y)^p}\\ &=[y^K]\,(1-y) \left(\frac{1}{1+\tfrac{y}{1-y}}\right)^{m+1}\\ &=[y^K]\,(1-y)^{m+2}\\ &=(-1)^K\,{m+2 \choose K},\,\mbox{ as desired }. \end{align*}