Euler Phi Function of Product with Prime

Theorem
Let $p$ be prime and $n \in \Z: n \ge 1$.

Then:
 * $\phi \left({p n}\right) = \begin{cases}

\left({p - 1}\right) \phi \left({n}\right) & : p \nmid n \\ p \phi \left({n}\right) & : p \mathrel \backslash n \end{cases}$ where $\phi \left({n}\right)$ denotes the Euler $\phi$ function of $n$.

Thus for all $n \ge 1$ and for any prime $p$, we have that $\phi \left({n}\right)$ divides $\phi \left({p n}\right)$.

Proof
First suppose that $p \nmid n$.

Then by Prime not Divisor implies Coprime, $p \perp n$.

So by Euler Phi Function is Multiplicative, $\phi \left({p n}\right) = \phi \left({p}\right) \phi \left({n}\right)$.

It follows from Euler Phi Function of Prime that $\phi \left({p n}\right) = \left({p - 1}\right) \phi \left({n}\right)$.

Now suppose that $p \mathrel \backslash n$.

Then $n = p^k m$ for some $k, m \in \Z: k, m \ge 1$ such that $p \perp m$.

Then:

At the same time: