Prove that $\phi (n^m) = n^{m−1} \phi (n)$.
Let's write $n$ as its prime decomposition
$$ n = p_1^{k_1} \times p_2^{k_2} \times \ldots \times p_z^{k_z} $$
where $p_i$ are primes, and $k_j$ are natural numbers. This means
$$ \begin{align} n^m & = p_1^{mk_1} \times p_2^{mk_2} \times \ldots \times p_z^{m k_z} \\ \\ n^{m-1} & = p_1^{(m-1)k_1} \times p_2^{(m-1)k_2} \times \ldots \times p_z^{(m-1) k_z} \end{align} $$
We have
$$ \phi (n) = p_1^{k_1-1}(p_1 - 1) \times p_2^{k_2-1}(p_2 - 1) \times \ldots \times p_z^{ k_z-1}(p_z - 1) $$
And so
$$ \begin{align} \phi(n^m) & = \phi(p_1^{mk_1}) \times \phi(p_2^{mk_2}) \times \ldots \times \phi(p_z^{m k_z}) \\ \\ & = p_1^{mk_1-1}(p_1-1) \times p_2^{mk_2 - 1}(p_2 - 1) \times \ldots \times p_z^{mk_z-1}(p_z - 1) \\ \\ & = \biggl( p_1^{(m-1)k_1} \times p_2^{(m-1)k_2} \times \ldots \times p_z^{(m-1) k_z} \biggr) \times \\ & \quad \; \biggl( p_1^{k_1-1}(p_1 - 1) \times p_2^{k_2-1}(p_2 - 1) \times \ldots \times p_z^{ k_z-1}(p_z - 1) \biggr) \\ \\ &= n^{m−1} \phi (n) \end{align} $$
The last line is justified by noting that $(m-1)k_i + k_i -1 = mk_i - 1$.
We've shown $\phi(n^m) = n^{m−1} \phi (n)$.