How would I find a formula for $$S(n,r) = \sum_{i_1+\ldots+i_r = n,~(i_k)\in\mathbb{N}^r} ~~~i_1\ldots i_r ~~~~.$$
It's easy to find that it satisfies $$ S(n,r+1) = \sum_{j=0}^n(n-j)S(j,r),$$ which shows that it is a polynomial in $n$ (at $r$ fixed). Also, $S(n,1) = n$, $S(n,2)=\frac{(n-1)n(n+1)}{6}$ (if my calculations are not wrong).
Edit: also, if I let $f_r(z) = \sum_{n=0}^\infty S(n,r)z^n$, I have something like $$f_{r+1}(z) = \frac{zf_r(z)}{(1-z)^2}.$$ Thus $$ f_r(z) = \frac{z^r}{(1-z)^{2r}}.$$ That doesn't seem very helpful since that function expansion is hard to compute, so that's that. Something like $$S(n,r) = n!f^{(n)}_r(0).$$