$ϕ(m · n) = \frac{ϕ(m) · ϕ(n) · gcd(m, n)}{ϕ(gcd(m, n))}$
Deduce from the previous identity that $ϕ(n^k) = n^{k−1}· ϕ(n)$
I know that \begin{eqnarray*} \phi(m) = m \prod_{p \mid m } (1-\frac{1}{p}). \\ \end{eqnarray*}
How can I make use of that? Any help please?
If what you are looking for is using the identity $(1)\ \phi(nm)=\frac{\phi(n)\phi(m)d}{\phi(d)}$ (where $d=g.c.d(n,m)$) to prove that $\phi(n^k)=n^{k-1}\phi(n)$ then one way to do it is by induction :
For $k=1$ the equality $\phi(n^k)=n^{k-1}\phi(n)$ holds.
Let $k\ge 1$ and suppose $\phi(n^k)=n^{k-1}\phi(n)\ (2)$. And let's prove that $\phi(n^{k+1})=n^k\phi(n)$.
By replacing $n$ by $n^k$ and $m$ by $n$ in $(1)$ we get $d=n$ and
$\phi(n^{k+1})=\frac{\phi(n^k)\phi(n)n}{\phi(n)}=n\phi(n^k)$ And by $(2)$ we get $\phi(n^{k+1})=n^k\phi(n)$ hence the result.
