The number-theoretic Möbius function $\mu(n)$ from prime factorisation

For a positive integer $n$, the number-theoretic Möbius function is

$\mu(n):=\begin{cases}0,&v_p(n)\ge2\text{ for some prime }p,\\(-1)^r,&n=p_0p_1\cdots p_{r-1}\text{ for distinct primes }p_i.\end{cases}$

The power $(-1)^r$ is the natural power in the multiplicative monoid of $\mathbb Z$ (Powers $g^{n}$: natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$, The integers form a commutative ring).

This definition is well posed. Canonical prime factorisation (For $n \ge 1$ and any injective list $p : r \to \mathbb{Z}$ of primes containing every prime divisor of $n$, one has $n = \prod_{i<r} p_i^{\,v_{p_i}(n)}$; the exponents are determined by $n$, and $v_q(n) = 0$ for every prime $q$ outside the list, The $p$-adic valuation $v_p(a)$ of a nonzero integer: the greatest $k \in \mathbb{N}$ with $p^{k} \mid a$) uniquely determines every exponent $v_p(n)$. If none exceeds $1$, the primes with exponent $1$ form a finite list whose length $r$ is invariant under reordering by the uniqueness clause of The fundamental theorem of arithmetic: every integer $n \ge 1$ is a product of primes, and the factorisation is unique up to order — if $\prod_{i<r} p_i = \prod_{j<s} q_j$ with every $p_i$ and $q_j$ prime, then $r = s$ and $q_i = p_{\pi(i)}$ for some $\pi \in \operatorname{Sym}(r)$. If some exponent exceeds $1$, the first clause applies independently of which such prime is noticed. For $n=1$ the prime list is empty, so

$\mu(1)=(-1)^0=1.$

Equivalently, $\mu(n)=0$ exactly when a prime square divides $n$; otherwise its sign records the parity of the number of distinct prime factors (For a prime $p$ and a nonzero integer $a$: $p^{v_p(a)} \mid a$ and $p^{v_p(a)+1} \nmid a$; $p^{k} \mid a$ holds exactly for $k \le v_p(a)$; $v_p(a) \ge 1$ exactly when $p \mid a$; $v_p(1) = v_p(-1) = 0$; and $v_p(p) = 1$).