We want to calculate $$\rho(k,n,m)=\text{res}_{w=1}\left(\frac{w^n}{1-w^m}\right)^k\frac{dw}{w}. $$ If $kn$ is divisible by $m$ then it seems that $\rho(k,n,m)=-\binom{-k}{kn/m-k}/m$. This is because in this case the residues at all $m$'th roots of unity are the same, and the sum of those residues is minus the residue at $\infty$, which is easily calculated by the substitution $w=t^{-1}$ and the binomial expansion of $(1-t^m)^{-k}$. I have checked this in Maple for a range of cases. I don't know if this method can be adapted to the case where $kn$ is not divisible by $m$.

One can also check experimentally that the denominator and numerator of $\rho(k,n,m)$ are large, but their factorisation only involves fairly small primes $p$, certainly with $p<knm$. This typically indicates that the function can be expressed in terms of binomial coefficients and factorials, rather than general polynomials.

We want to calculate $$\rho(k,n,m)=\operatorname*{res}_{w=1}\left(\frac{w^n}{1-w^m}\right)^k\frac{dw}{w}. $$ If $kn$ is divisible by $m$ then it seems that $\rho(k,n,m)=-\binom{-k}{kn/m-k}/m$. This is because in this case the residues at all $m$'th roots of unity are the same, and the sum of those residues is minus the residue at $\infty$, which is easily calculated by the substitution $w=t^{-1}$ and the binomial expansion of $(1-t^m)^{-k}$. I have checked this in Maple for a range of cases. I don't know if this method can be adapted to the case where $kn$ is not divisible by $m$.

One can also check experimentally that the denominator and numerator of $\rho(k,n,m)$ are large, but their factorisation only involves fairly small primes $p$, certainly with $p<knm$. This typically indicates that the function can be expressed in terms of binomial coefficients and factorials, rather than general polynomials.