There is a very simple proof for Möbius inversion formula through convolution:
If $A$ is a UFD and $B$ is a ring, $f,g:A\rightarrow B$ two functions, then
$(f\ast g)(n) = \sum_{k\cdot l = n}f(k)\cdot g(l)$ makes the set of functions into a ring with identity $\delta_1$.
In this case the Möbius function, assigning 0 to every $n\in A$ which is not square free, 1 to square free elements which are the products of an even number of primes and -1 to square free products of odd number of primes, is an inverse of the constant function 1. Then the Möbius inversion formula
$f(n) = \sum_{d|n} g(d)\Longrightarrow g(n) = \sum_{d|n}f(d)\mu\left(\frac{n}{d}\right)$
is simply another way of writing
$f = g\ast 1\Longrightarrow g = f\ast \mu.$
Is there a similarly general and elegant approach to the multiplicative formula
$f(n) = \prod_{d|n}g(d)\Longrightarrow g(n) = \prod_{d|n} f(d)^{\mu(n/d)}?$
Thank you.