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: 
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.
I'm going to provide a secondary generating function which is probably more relevant to your updated question. Let's recall your definition,
$$
f(Q):=\sum\_{\substack{\sum\_{i=0}^k n\_i=n \\ n\_0,\ldots,n\_k \ \geq0 \\\sum\_{j=0}^kjn\_j=Q}} \binom{n}{n\_0,\cdots,n\_k}x\_{0}^{n\_0}\ldots x\_{k}^{n\_k}
$$
and let us calculate $
G(X):=\sum\_{Q=0}^{nk} f(Q)X^Q
$. Before, we do that note that for any $p\geq 0$, $$\boxed{\sum\_{Q=0}^{nk} Q^pf(Q)=\left.\left(X\frac{d}{dX}\right)^p G(X)\right|\_{X=1}}$$ 

First of all, note that under the condition $\sum\_{j=0}^k n\_j = n$ and $n\_0, \cdots, n\_k\geq 0$, one has
$$
nk -\sum\_{j=0}^k jn\_j=
\sum\_{j=0}^k (k-j) n\_j\geq 0
$$
As a result, the equation $z-\sum\_{j=0}^k jn\_j=0$ has exactly one solution for $z$ in the integer range $0\leq z\leq nk$. Using this, we have
$$
\begin{aligned}
G(X)&=\sum\_{Q=0}^{nk} X^{Q}
\sum\_{\substack{\sum\_{i=0}^k n\_i=n \\ n\_0,\ldots,n\_k \ \geq 0}} \binom{n}{n\_0,\cdots,n\_k}x\_{0}^{n\_0}\ldots x\_{k}^{n\_k}
\delta\left(Q-\sum\_{j=0}^kjn\_j\right)
\\
&=
\sum\_{\substack{\sum\_{i=0}^k n\_i=n \\ n\_0,\ldots,n\_k \ \geq 0}} \binom{n}{n\_0,\cdots,n\_k}\prod\_{m=0}^k (x\_mX^m)^{n\_m}=\left(\sum\_{m=0}^k x\_m X^m\right)^{n}
\end{aligned}
$$
So to summarize:
$$
\boxed{
G(X)=\sum\_{Q=0}^{nk} f(Q) X^Q = \left(\sum\_{m=0}^k x\_m X^m\right)^{n}
}
$$
This should make the rest obvious.