Prove that $ \frac{n}{\phi(n)} = \sum\limits_{d \mid n} \frac{\mu^2(d)}{\phi(d)} $
$$\sum_{d|n}\frac{\mu^2(d)}{\phi(d)}=1+\frac{1}{p_1-1}+\frac{1}{p_2-1}+\cdots+\frac{1}{(p_1-1)(p_2-1)+\cdots}$$
$$=(1+\frac{1}{p_1-1})(1+\frac{1}{p_2-1})\cdots=\prod_{p_i|n}(1+\frac{1}{p_i-1})=\prod_{p_i|n}(\frac{1}{1-1/p_i}) $$
$$ =\frac{n}{\phi(n)} $$
Prove both sides (of the original equation) are multiplicative, then prove they're equal for prime powers.