Prove that
$$ \phi ( \phi (p^k)) = p^{k−2} \phi [p(p− 1)] $$
where $p$ is prime and $k \ge 2$.
We proceed as follows
$$ \phi(p^k) = p^{k-1}(p-1)$$
Since $p^{k-1}$ and $(p-1)$ are coprime, we can use the multiplicity of $\phi$
$$ \phi(\phi(p^k)) = p^{k-2}(p-1) \phi (p-1) $$
Similarly, $p$ and $(p-1)$ are coprime, we can again use the multiplicity of $\phi$
$$ \phi(\phi(p^k)) = p^{k-2} \phi [p(p-1)] $$