Multiplicative Möbius inversion formula. 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.
 A: Of course there is:
$$\tag1 \ln f=\ln g *1\Rightarrow \ln g=\ln f *\mu.$$
At least this works if we can take logarithms in $B$.
Otherwise, we need to switch to a different ring:
Assume $B$ is a commutative ring with unity. Let $C$ be the set of permutations maps $\sigma \colon B^\times\to B^\times$ of the units with $\sigma(1)=1$.
Then $C$ is a ring with pointwise multiplication as addition, with composition as multiplication, identity as one and $x\mapsto 1$ as zero. Moreover, we can map $\mathbb Z\to C$ via $n\mapsto(x\mapsto x^n)$ and consider $B^\times$ as a subset of $C$ via $b\mapsto(x\mapsto b x)$.
Now if $f,g\colon A\to B$ are two functions with range actually in $B^\times\subseteq C$, then $f=g*1\Rightarrow g=f*\mu$ is the multiplicative inversion formula
$$\tag2 f(n)=\prod_{d|n}g(d)\Rightarrow g(n)=\prod_{d|n}f(d)^{\mu(\frac nd)}.$$
What can be done if the ranges of $f,g$ are not contained in $B^\times$? Then let us hope that $B$ is at least an integral domain and replace $B$ with its quotient field.
Your specific example is with polynomials, i.e. the integral domain $\mathbb Z[X]$, so the above steps work fine.
In that specific case, maybe the following is more intuitive: 
We can interprete polynomials as functions $\mathbb C\to \mathbb C$. Outside their roots, we can take (multivalued) logarithm, thus can apply $(1)$ pointwise and obtain the conclusion in $(2)$ pointwise for all but finitely many points $\in\mathbb C$, hence $(2)$ also as a whole.
