alternate proof of Möbius inversion formula

The Möbius inversion theorem can also be proved elegantly using the fact that arithmetic functions form a ring under $+$ and $*$.

Let $I$ be the arithmetic function that is everywhere $1$. Then obviously if $\mu$ is the Möbius function,

and thus $I*\mu =e$, where $e$ is the identity of the ring.

But then

$$f(n)=\sum _{d|n}g(d)=\sum _{d|n}g(d)I\left(\frac{n}{d}\right)$$

and so $f=g*I$. Thus $f*\mu =g*I*\mu =g$. But $g=f*\mu$ means precisely that

$$g(n)=\sum _{d|n}\mu (d)f\left(\frac{n}{d}\right)$$

and we are done.

The reverse equivalence is similar ($f*\mu =g\Rightarrow f*\mu *I=g*I\Rightarrow f=g*I$).