I have to demonstrate that $$\phi^{-1}(n)= \prod_{p|n}(1-p)$$ where $\phi(n)$ is the Euler's totient function.
I know that I can write $\phi$ in terms of the Mobius function $\mu$ as$$\phi(n)= \sum_{n|d} \mu(d) \cdot \left(\frac{n}{d}\right)$$ and I tried to substitute this in the definition of Dirichlet's inverse $$f^{-1}(n)=- \frac{1}{f(1)} \sum_{d|n \\d<n} f \left(\frac{n}{d}\right)f^{-1}(d) $$ but I can't find a solution.
Can anyone give me a hint?
Since $\phi(n)$ is multiplicative, i.e., satisfies $\phi(nm) = \phi(n)\phi(m)$ for all $m,n$ with $gcd(n,m) = 1$, it follows from $\phi(p^k)= p^{k-1}(p-1)$ that $$\sum_{d | n} \phi(d) = n, \qquad \phi(n) = \sum_{d | n} \mu(d) \frac{n}{d}$$ by the Moebius inversion formula. Thus $$\phi^{-1}(n) = \sum_{d | n} d \mu(d)$$ by the definition of the Dirichlet inverse. But we have, for every multiplicative arithmetic function $f$ that $$ \sum_{d\mid n}\mu(d)f(d)=\prod_{p\mid n}(1-f(p)). $$ Just check this for prime powers and use the multiplicativity of $f$. Now take $f=id$ to obtain $$\phi^{-1}(n)= \prod_{p|n}(1-p).$$