This question is question 2.3 of Apostol Introduction to analytic number theory.
It's solution image:
My question: how did in RHS on last line of solution author got $n / \phi(n) $ as $n = p_{1}^{a_1} ... p_{k}^{a_k}$ not $ p_{1}... p_{k} $ ? and higher powers of p becomes 0 due to mobius function.
So, why $ n / \phi(n) $ ?
The powers of the primes does not matter because they will get canceled, $n= p_1^{e_1} \cdots p_k^{a_k}$ and so we have that $\phi(n) = \phi(p_1^{a_1}) \cdots \phi(p_k^{a_k}) = p_1^{a_1-1} (p_1-1) * p_2^{a_2-1} (p_2-1) \cdots p_k^{a_k-1} (p_k-1)$, this formula due to the fact that $\phi(p^k) = p^{k-1} (p-1)$ for primes and that $\phi$ is multiplicative function.
And we get that $ \frac{n}{\phi(n)} = \frac{p_1^{a_1}}{p_1^{a_1-1} (p_1-1)} \cdots \frac{p_k^{a_k}}{p_k^{a_k-1}(p_k-1)} = \prod \limits_{i=1}^{k} \frac{p_i}{p_i-1} = \prod \limits_{i=1}^{k} \frac{1}{1-\frac{1}{p_i}}$
I hope this clarify the last line of the solution.