The question:
Let $(a_n)_{n=1}^k\subset \mathbb{N}$ be positive. Find the component $c_{-k}$ in Laurent expansion of $$R(z)=\dfrac{1}{((z^{a_1}-1)\cdots(z^{a_k}-1))}$$ in the neighbourhood of $z=1$.
My attampt:
$$\begin{align} c_{-k}&= {1\over 2\pi i}\oint_{C_R}\dfrac{(z-1)^{k-1}}{(z^{a_1}-1)\cdots(z^{a_k}-1)}dz\\&=\sum Res(g(z):=\dfrac{(z-1)^{k-1}}{(z^{a_1}-1)\cdots(z^{a_k}-1)},a_{l,j}) \end{align}$$ where $$a_{l,j}=\exp({2\pi ji\over l})$$ for all $l=a_n$ for some $n$ and $0\leq j<l$.
This could be a final solution but I wonder whether there's more elgant solution.
Well, $$f(z)=(z-1)^k R(z)=\frac1{P_{a_1}(z)P_{a_2}(z)\cdots P_{a_k}(z)}$$ where $$P_a(z)=1+z+z^2+\cdots+z^{a-1}.$$ Therefore $f$ is holomorphic in a neighbourhood of $z=1$, and $$c_{-k}=f(1)=\frac1{a_1a_2\cdots a_k}.$$