Looking for a proof for the value of the residue. Trying to simplify a calculation I have found by numerical experiments the following interesting result:
$$
\underset{z=1}{\operatorname{Res}}\frac{z^{p-1}}{(z^n-1)^q}=
\frac qp\frac{\left(\frac pn\right)^{\underline{q}}}{q!}
\equiv\frac qp\binom{\frac pn}q,
$$
where $p,q,n$ are positive integers and $x^{\underline r}=x(x-1)\cdots(x-r+1)$ means the falling factorial.
Is there a simple way to prove this?
 A: 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 
$\mathop{\rm Res}_{z=1}\frac{z^{p-1}}{z^n-1}=\frac{z^{p-1}}{(z^n-1)'}\bigg|_{z=1}=\frac1n$.)
A: The next one-line derivation is based on the following lemma:
$$
\underset{z=h(a)}{\operatorname{Res}}f(z)=\underset{z=a}{\operatorname{Res}}f(h(z))h'(z),\tag1
$$
where $h$ is analytic in the neighborhood of $a$ and $h'(a)\ne0$. 
Let
$$
f(z)=\frac{z^{p-1}}{(z^n-1)^q};\quad h(z)=(1+z)^\frac1n.
$$
Observe that $h(z)$ is analytic in the neighborhood of $z=0$, $h(0)=1$ and $h'(0)=\dfrac1n\ne0$.
We have:
$$
f(h(z))h'(z)=\frac{(1+z)^\frac{p-1}n}{z^q}\frac{(1+z)^{\frac1n-1}}n
=\frac{(1+z)^{\frac pn-1}}{nz^q}\stackrel{(|z|<1)}=\frac{\sum_{k\ge0}\binom{\frac pn-1}kz^k}{nz^q}\\
\implies \underset{z=1}{\operatorname{Res}}f(z)=
\underset{z=0}{\operatorname{Res}}f(h(z))h'(z)=\frac{1}{n}\binom{\frac pn-1}{q-1}
=\frac qp\binom{\frac pn}{q}.
$$
