Looking for a proof for the value of the residue.

For $k\ge1$, consider $$ \left(\frac{z^p}{(z^n-1)^k}\right)' = (p-nk)\frac{z^{p-1}}{(z^n-1)^k} - nk\frac{z^{p-1}}{(z^n-1)^{k+1}}. $$ The LHS is a derivative, so its residue is $0$ anywhere. Hence, $$ \mathop{\rm Res}_{z=1}\frac{z^{p-1}}{(z^n-1)^{k+1}} = \frac{\frac{p}{n}-k}{k} \cdot \mathop{\rm Res}_{z=1}\frac{z^{p-1}}{(z^n-1)^k}. $$ Repeating this step for $k=q-1,q-2,\ldots,1$ we obtain $$ \mathop{\rm Res}_{z=1}\frac{z^{p-1}}{(z^n-1)^q} = \left(\prod_{k=1}^{q-1} \frac{\frac{p}{n}-k}{k}\right) \cdot \mathop{\rm Res}_{z=1}\frac{z^{p-1}}{z^n-1} = \binom{\frac{p}{n}-1}{q-1} \cdot \frac1n = \frac{q}{p} \cdot \binom{\frac{p}{n}}{q}. $$ (In the last step, in $\frac{z^{p-1}}{z^n-1}$ the denominator has a single root at $1$, so $\mathop{\rm Res}_{z=1}\frac{z^{p-1}}{z^n-1}=\frac{z^{p-1}}{(z^n-1)'}\bigg|_{z=1}=\frac1n$.)