Question: 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.

Answer: 
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, $(Q!)^{-1}\partial\_X^n \partial\_Y^Q F\mid\_{X=Y=0}$).
Starting from Hamed's answer, we can give a quick proof that $\sum\_{Q\ge 0} Qf(Q)$ equals what you think it is. They proved
$$
\sum\_{n\ge 0}\sum\_{q\ge 0}\frac1{n!}f\_n(Q)X^nY^q = \exp\Big(\sum\_{m=0}^k x\_mXY^m\Big)
$$
Differentiating both sides with respect to $Y$, and then multiplying by $Y$, you get
$$
\sum\_{n\ge 0}\sum\_{q\ge 0}\frac1{n!}Qf\_n(Q)X^nY^q = \big(\sum\_{k=0}^m mx\_m XY^m\big)\exp\Big(\sum\_{m=0}^k x\_mXY^m\Big)
$$
Notice that we now have a $Qf\_n(Q)$. These need to be summed, so set $Y=1$:
$$
\sum\_{n\ge 0}\sum\_{q\ge 0}\frac1{n!}Qf\_n(Q)X^n = \big(\sum\_{k=0}^m mx\_m X\big)\exp\Big(X\sum\_{m=0}^k x\_m\Big)
$$
Finally, extract the coefficient of $X^n$ of both sides. On the left hand side, $X\_n$ has a $\frac1{n!}\sum\_{q\ge 0}QF\_n(Q)$ attached to it. On the right, you get the $X^n$ coefficient by multiplying $\big(\sum\_{k=0}^m mx\_m\big)$ with the $X^{n-1}$ coefficient of $\exp\Big(X\sum\_{m=0}^k x\_m\Big)$. This implies
$$
\frac1{n!}\sum\_{q\ge 0}QF\_n(Q)=\big(\sum\_{k=0}^m mx\_m\big)\cdot \frac1{(n-1)!}\Big(\sum\_{k=0}^m x\_m\Big)^{n-1}
$$
which is what you wanted.