I came across this question regarding Möbius Inversion. The question was "How can I show that for all $n \in \mathbb{N}$;$$\sum_{d \mid n} \mu(d)\sigma(n/d) = n$$ Where $\sigma$ is the divisor function and $\mu$ is the Möbius function." The answer was "$$\sum_{d \mid n} d = \sigma(n)$$ and so by the Möbius inversion formula, we have that
$$\sum_{d \mid n} \mu(d) \sigma\left(\frac{n}{d}\right) = n".$$
My question is, what if the arguments were swapped? That is, how would you prove that if $f(n)=\sum_{d|n}\sigma(d)\mu(n/d) $ then $f(n)=n$?
Some properties of the "divisor function":
$\displaystyle \sigma (n) = \sum_{d|n} d$.
$\displaystyle \sigma (p^e) = 1+p+p^2+...+p^e = \frac{p^{e+1}-1}{p-1}$.
$\sigma$ is multiplicative.