Find the coefficients of negative powers of Laurent expansion of $R(z)={1\over (z^{a_i}-1)\dots(z^{a_n}-1)}$ about the point $z=1$, where $a_i\in\mathbb{N}$.
I think I have an answer but I would like to verify it:
For all $|z|<1$, and for all $i$, $${1\over z^{a_i}-1}={-1\over 1-z^{a_i}}\\=-\sum z^{k\cdot a_i}$$ Thus, $$f(z)=\prod_{i=1}^n(-\sum_{k=0}^\infty z^{k\cdot a_i})$$ Thus, for any $\mathbb{Z}\ni l<0$, the coeffitient $c_l$ in the Laurent epansion of $f$ equals $0$.
This expansion will let you constract the Laurent series around $z=0$, and for this series indeed the coefficients with negative indexes are equal to $0$, but this is not what you're looking for.
To get the Laurent series around $z=1$ you need to use $$ \frac{1}{z^{a}-1}= \frac{1}{\big(1+(z-1)\big)^a-1} = \frac{1}{\sum_{k=1}^a \binom{a} {k}(z-1)^k} = \frac{1}{a(z-1)} \frac{1}{1+\frac1a\sum_{k=2}^a \binom{a} {k}(z-1)^{k-1}} = \\ = \frac{1}{a(z-1)} \sum_{m=0}^\infty \frac{(-1)^m}{a^{m}}\Big(\sum_{k=1}^{a-1} \binom{a}{k+1}(z-1)^k\Big)^m = \\ = \frac{1}{a(z-1)} \Bigg(1+\sum_{m=1}^\infty \frac{(-1)^m}{a^m}\sum_{\begin{array}{c}k_1,\dots k_m\\1\le k_i \le a-1\end{array}}\binom{a}{k_1+1}\dots \binom{a}{k_m+1}(z-1)^{k_1+\dots k_m}\Bigg) = \\ =\frac{1}{a(z-1)} \Bigg(1+\sum_{m=1}^\infty \frac{(-1)^m}{a^m}\sum_{N=m}^\infty \sum_{\begin{array}{c}k_1,\dots k_m\\1\le k_i \le a-1 \\ k_1+\dots+k_m=N\end{array}}\binom{a}{k_1+1}\dots \binom{a}{k_m+1}(z-1)^N\Bigg) = \\ = \frac{1}{a(z-1)} \Bigg(1+\sum_{N=1}^\infty \sum_{m=1}^N\frac{(-1)^m}{a^m} \sum_{\begin{array}{c}k_1,\dots k_m\\1\le k_i \le a-1 \\ k_1+\dots+k_m=N\end{array}}\binom{a}{k_1+1}\dots \binom{a}{k_m+1} (z-1)^N\Bigg) = \\ = \frac{1}{a(z-1)} \Bigg(1+\sum_{N=0}^\infty \sum_{m=1}^{N+1}\frac{(-1)^m}{a^m} \sum_{\begin{array}{c}k_1,\dots k_m\\1\le k_i \le a-1 \\ k_1+\dots+k_m=N+1\end{array}}\binom{a}{k_1+1}\dots\binom{a}{k_m+1} (z-1)^{N+1}\Bigg)$$ that is $$ \frac{1}{z^a-1} = \sum_{N=-1}^\infty c_N(a) (z-1)^N$$ where $$ c_{-1}(a) =\frac 1 a$$ $$ c_N(a) = \sum_{m=1}^{N+1}\frac{(-1)^m}{a^{m+1}} \sum_{\begin{array}{c}k_1,\dots k_m\\1\le k_i \le a-1 \\ k_1+\dots+k_m=N+1\end{array}}\binom{a}{k_1+1}\dots \binom{a}{k_m+1} \qquad \text{for }N\ge 0$$ This is a fairly complicated formula, but it can be evaluated to any order. Multiplying such formulas will give us the Laurent expansion of $\frac{1}{(z^{a_1}-1)\dots(z^{a_n}-1)}$. It's easy to see the first non-vanishing term in the Laurent series; it will come from myultiplying terms $\frac{1}{a}(z-1)^{-1}$ and it will be $$ \frac{1}{a_1\dots a_n}(z-1)^{-n}$$ To calculate the subsequent terms you'll need values of $c_N(a_i)$ although you can notice that you'll only need them for $N<n-1$. If you'd take one term with the power $(z-1)^{n-1}$, then since the remaining terms will all have powers $(z-1)^{-1}$ or higher, althogether their product will have power $(z-1)^0$ or higher, i.e. it will not contribute to the negative part of the Laurent series.