Show that $\frac1 {\phi(n)} = \frac1 n \sum_{d|n} \frac{\mu(d)^2}{\phi(d)}$ for all positive integers n
In class we solved this question by observing that both sides are multiplicative functions and they should be equal at prime values. There is this another method we have done that I cant understand. I have to express right-hand side(the summation) as a multiplication but I just cant figure out. It would be better if you could give me a hint. Thanks in advance.
Suppose we write this as
$$\frac{n}{\varphi(n)} = \prod_{d|n} \frac{\mu(d)^2}{\varphi(d)}.$$
The LHS is by definition
$$\prod_{p|n} \frac{1}{1-\frac{1}{p}} = \prod_{p|n} \frac{p}{p-1}$$
where the product is over prime divisors of $n.$ Now since $\mu(n)$ and $\varphi(n)$ are multiplicative and the product on the RHS only includes terms $d$ that are products of primes we get for the RHS
$$\prod_{p|n} \left(1+\frac{\mu(p)^2}{\varphi(p)}\right) = \prod_{p|n} \left(1+\frac{1}{p-1}\right) = \prod_{p|n} \frac{p}{p-1}$$
the same as the LHS and we are done.