Is there an explicit formula for the following quantity?


$$f\_m(a\_1,\ldots,a\_n):=\sum\_{\substack{k\_1+\ldots+k\_n=m \\ k\_1,\ldots,k\_n\in \mathbb{N}}} k\_1^{a\_1}\ldots k\_n^{a\_n}\ ,\hspace{1cm} m,a\_1,\ldots,a\_n\in \mathbb{N}$$


(for instance $f\_m(0,\ldots,0)$ is simply the number of compositions of m into n parts, $f\_m(1,1)=\frac{m(m+1)(m-1)}{6}$ and so on). I would like an answer both for the case where the $k\_i$'s can and cannot attain the value $0$, if possible.



Sum over integer compositions of $m$ with $n$ parts of a fixed monomial in the parts
I'm not sure if there is anything simpler than $f\_m(a\_1,\dots,a\_n)$ being the coefficient of $x^m$ in the product of [polylogarithms](http://en.wikipedia.org/wiki/Polylogarithm):
$$\mathrm{Li}\_{-a\_1}(x)\cdot \mathrm{Li}\_{-a\_2}(x)\cdots \mathrm{Li}\_{-a\_n}(x).$$


In particular, since $\mathrm{Li}\_{-1}(x) = x\cdot (1-x)^{-2}$, we have
$$f\_m(\underbrace{1,1,\dots,1}\_n) = \mathrm{Coeff}\_{x^m} x^n (1-x)^{-2n} = (-1)^{m-n}\cdot \binom{-2n}{m-n} = \binom{m+n-1}{2n-1}.$$


