Show that $\phi (p^m) = \phi (p) p^{m−1}$ where $p$ is a prime and $m$ is a natural number.
We use Proposition (5.4) which says that if $p$ is prime, then $\phi(p^k) = p^{k-1}(p-1)$ where $k$ is a natural number.
We also remember that $\phi(p)=p-1$.
And so
$$ \phi(p^m) = p^{m-1}(p-1) = \phi(p) p^{m-1}$$