Cyclotomic Polynomial of Index times Prime Power

Theorem
Let $n, k \ge 1$ be natural numbers.

Let $p$ be a prime number.

Let $\Phi_n$ denote the $n$th cyclotomic polynomial.

Then $\map {\Phi_{p^k n}} x = \begin{cases} \map {\Phi_n} {x^{p^k}} & \text{if } p \divides n\\ \dfrac {\map {\Phi_n} {x^{p^k}}} {\map {\Phi_n} {x^{p^{k - 1}}}} & \text{if } p \nmid n \end{cases}$

Proof
Suppose $p \divides n$.

Then for all $m \in \Z$:

Hence:

Now suppose $p \nmid n$.

We still have $p \divides p n$.

Write $p^k = p^{k - 1} p n$.

Notice that the result we proved above holds trivially for $k = 0$:
 * $\map {\Phi_{p^0 n} } x = \map {\Phi_n } x = \map {\Phi_n } {x^1} = \map {\Phi_n } {x^{p^0}}$

Hence from the above:
 * $\map {\Phi_{p^k n} } x = \map {\Phi_{p n}} {x^{p^{k - 1}}}$

We need the following result:
 * the sets $\set {m \in \Z: m \perp p n}$ and $\set {p r: r \perp n}$ are disjoint and has union $\set {m \in \Z: m \perp n}$

First to show that they are indeed disjoint:

Suppose $x \in \set {p r: r \perp n}$.

Then $p \divides x$.

Since $p \divides p n$:
 * $x \not \perp p n$

and thus:
 * $x \notin \set {m \in \Z: m \perp p n}$

Hence the sets are disjoint.

Now we show that their union is indeed $\set {m \in \Z: m \perp n}$.

By Divisor of One of Coprime Numbers is Coprime to Other:
 * $\forall m \in \Z: m \perp p n \implies \paren {m \perp p \land m \perp n}$

This gives:
 * $\set {m \in \Z: m \perp p n} \subseteq \set {m \in \Z: m \perp n}$

Let $x \in \set {p r: r \perp n}$.

We are given that $p \perp n$.

By Integer Coprime to all Factors is Coprime to Whole:
 * $x \perp n$

Hence $x \in \set {m \in \Z: m \perp n}$.

This gives:
 * $\set {p r: r \perp n} \subseteq \set {m \in \Z: m \perp n}$

By Union of Subsets is Subset:
 * $\set {m \in \Z: m \perp p n} \cup \set {p r: r \perp n} \subseteq \set {m \in \Z: m \perp n}$

For the other direction, we let $x \notin \set {m \in \Z: m \perp p n} \cup \set {p r: r \perp n}$.

Then by De Morgan's Laws (Set Theory)/Set Complement:
 * $x \in \set {m \in \Z: m \not \perp p n} \cap \set {p r: r \not \perp n}$.

By definition of intersection:
 * $x \in \set {p r: r \not \perp n}$

Thus:
 * $\exists d \in \Z: d > 1: d \divides r \divides x \land d \divides n$

Therefore $x \not \perp n$.

This gives:
 * $x \notin \set {m \in \Z: m \perp n}$

Hence:
 * $\set {m \in \Z: m \perp n} \subseteq \set {m \in \Z: m \perp p n} \cup \set {p r: r \perp n}$

and we have our result by definition of set equality.

Therefore:

as required.