I am currently learning about Möbius Inversion. In Wikipedia , the sum version is proved using convolutions which I can follow along with. However, it also lists a product version of this formula:
$g(n)= \Pi_{d|n} f(d)$ iff $f(n)=\Pi_{d|n} g(\frac{n}{d})^{\mu(d)}$
I want to know how you can prove this statement. I cannot see how I could use convolutions for this since they involve sums and this one needs products. Is there some theorem/technique I should be utilizing for this?
One can try to rewrite $\prod_{d|n} g(\frac{n}{d})^{\mu(d)}$,
$\prod_{d|n} g(\frac{n}{d})^{\mu(d)}=\prod_{d|n} (\prod_{d'|\frac{n}{d}} f(d'))^{\mu(d)}\\ =\prod_{d'|n}\prod_{d|\frac{n}{d'}}f(d')^{\mu(d)}=\prod_{d'|n}f(d')^{\sum_{d|\frac{n}{d'}}\mu(d) }$.
But $\sum_{d|\frac{n}{d'}}\mu(d)=0$ if $\frac{n}{d'}\not=1$ and $=1$ otherwise.
Thus $\prod_{d|n} g(\frac{n}{d})^{\mu(d)}=f(n)^1=f(n)$.