# Expression for the inverse of Euler's totient function $\phi^{-1}$

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 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 $$\phi^{-1}(n)= \prod_{p|n}(1-p).$$