Prove $\sum_{d|n} \frac{\Phi(d)}{d} = \prod_{i=1}^r (1 + a_i - \frac{a_i}{p_i})$ I want to prove $\sum_{d|n} \frac{\Phi(d)}{d} = \prod_{i=1}^r (1 + a_i(1 - \frac{1}{p_i}))$, where $\Phi(n)$ is the Euler phi function and given the prime factorisation $n = \prod_{i=1}^r p_i^{a_i} $.
My instinct says to use the Möbius inversion formula, but I am struggling to identify the functions $f(n)$ and $g(n)$ that will allow me to prove this. Any insight would be appreciated.

Assuming that your
$\Phi$
is actually the
Euler $\phi$ function,
use the fact that,
since
$\phi$ is a multiplicative
function,
so is
$\sum_{d | n}\dfrac{\phi(d)}{d}$.
Then you only need to prove it for
$p^a$.
