Prove that
$$ \phi (p^kq^k) = p^{k−1}q^{k−1} \phi (p) \phi (q) $$
where $p$ and $q$ are distinct primes.
We have
$$ \begin{align} phi(p^kq^k) & = \phi(p^k) \times \phi(q^k) \\ \\ & = p^{k-1}(p-1) \times q^{k-1}(q-1) \\ \\ & = p^{k−1}q^{k−1} \phi (p) \phi (q) \end{align} $$
The last step uses $\phi(p)=p-1$.