Proof of the multiplicativity of the Möbius function

Proof

1. No prime square divides $1$, so $1$ is square-free. No prime divides $1$, so $\omega(1) = 0$, where $\omega$ is the number of distinct prime factors. By the definition of the Möbius function, $\mu(1) = (-1)^0 = 1$.
2. Let $m$ and $n$ be coprime positive integers.
3. If $m$ is not square-free, then $q^2 \mid m$ for some prime $q$, so $q^2 \mid mn$. Then $\mu(m) = 0$ and $\mu(mn) = 0$, so $\mu(mn) = \mu(m)\mu(n)$. The case where $n$ is not square-free is the same.
4. Now let $m$ and $n$ be square-free. Suppose $q^2 \mid mn$ for a prime $q$. By Euclid's lemma, $q \mid m$ or $q \mid n$, and not both, because $\gcd(m, n) = 1$. Say $q \mid m$ and $q \nmid n$.
5. With the p-adic valuation and the valuation of a product, $\nu_q(m) = \nu_q(m) + \nu_q(n) = \nu_q(mn) \ge 2$. So $q^2 \mid m$, a contradiction. Hence $mn$ is square-free.
6. A prime divides $mn$ if and only if it divides $m$ or $n$, by Euclid's lemma. No prime divides both, since $\gcd(m, n) = 1$. So $\omega(mn) = \omega(m) + \omega(n)$.
7. Therefore $\mu(mn) = (-1)^{\omega(m) + \omega(n)} = (-1)^{\omega(m)} (-1)^{\omega(n)} = \mu(m) \mu(n)$.
8. With step 1, $\mu$ is a multiplicative function.