Recall the Möbius inversion formula $1*\mu=\varepsilon$, where $\varepsilon$ denotes the convolution identity function. Thus, $f(1*\mu)=f\varepsilon$. Since pointwise multiplication of a completely multiplicative function distributes over convolution, $(f\cdot 1)*(f\mu)=f\varepsilon$. Note that, for all natural numbers $n$, $f(n)1(n)=f(n)\cdot 1=f(n)$ and $f(n)\varepsilon(n)=\varepsilon(n)$. Thus, $f*(f\mu)=\varepsilon$. It follows that $f\mu$ is the convolution inverse of $f$. ∎