Multinomial Expansion Given $x_0 \ldots x_k$ and $n$, Define $$f(Q)=\sum_{\substack{n_0+\ldots+n_k=n \\ n_0,\ldots,n_k \ >=0 \\n_1+2*n_2+\ldots+k*n_k=Q}} \binom{n}{n_0,\cdots,n_k}x_{0}^{n_0}\ldots x_{k}^{n_k}$$. Note that $$\sum_{Q=0}^{n*k} f(Q)=(\sum_{i=0}^{k}x_i)^n$$ which comes from the multinomial expansion. I was wondering how to calculate $\sum_{Q=0}^{n*k} Q\cdot f(Q)$.
I've checked several small cases, and it seems that $\sum_{Q=0}^{n*k} Q\cdot f(Q) = n(\sum_{i=0}^{k}x_i)^{n-1}(\sum_{i=0}^{k}ix_i)$, but I'm not sure how to prove that.

You can calculate the generating function instead. Define
$$
\begin{aligned}
F(X,Y)&=
\sum_{n=0}^\infty X^n\sum_{Q=0}^\infty Y^Q
\sum_{\substack{n_0+\ldots+n_k=n \\ n_0,\ldots,n_k \ >=0 \\n_1+2*n_2+\ldots+k*n_k=Q}} \frac{1}{n_0!n_1!\cdots n_k!}x_{0}^{n_0}\ldots x_{k}^{n_k}\\
&=\sum_{n=0}^\infty X^n\sum_{Q=0}^\infty Y^Q
\sum_{n_0, \cdots, n_k\geq 0} \frac{1}{n_0!n_1!\cdots n_k!}x_{0}^{n_0}\ldots x_{k}^{n_k}
\delta\left(n-\sum_{j=0}^k n_j\right)
\delta\left(Q-\sum_{j=0}^k jn_j\right)\\
&=
\sum_{n_0, \cdots, n_k\geq 0} 
\frac{1}{n_0!n_1!\cdots n_k!}x_{0}^{n_0}\ldots x_{k}^{n_k} X^{\sum_j n_j} Y^{\sum_j jn_j}\\
&=\prod_{m=0}^k \exp(x_m XY^m)
\end{aligned}
$$
where by $\delta(Z)$ I mean the Kronecker delta equating $Z=0$. What you are looking for the $n!$ times the coefficient of $X^nY^Q$ in the expansion of $F(X,Y)$ (in other words, $(Q!)^{-1}\partial_X^n \partial_Y^Q F\mid_{X=Y=0}$).
