Proof that the Möbius function is the inverse of the constant function 1

Proof

1. Let $n$ be a positive integer. By the definition of the Dirichlet convolution, $(\mu * \mathbf{1})(n) = \sum_{d \mid n} \mu(d) \mathbf{1}(n/d) = \sum_{d \mid n} \mu(d)$.
2. Case $n = 1$. The only divisor is $d = 1$. No prime square divides $1$, so $1$ is squarefree, and $\omega(1) = 0$. So the sum is $\mu(1) = (-1)^0 = 1$.
3. Case $n > 1$. By the fundamental theorem of arithmetic, $n = p_1^{e_1} \cdots p_k^{e_k}$ with distinct primes $p_i$, exponents $e_i \ge 1$, and $k \ge 1$.
4. By the definition of the Möbius function, $\mu(d) = 0$ unless $d$ is squarefree. So only squarefree divisors $d$ of $n$ contribute to the sum.
5. A divisor of $n$ has the form $p_1^{c_1} \cdots p_k^{c_k}$ with $0 \le c_i \le e_i$. It is squarefree exactly when each $c_i$ is $0$ or $1$.
6. So the squarefree divisors of $n$ are the products $d_S = \prod_{i \in S} p_i$, one for each subset $S$ of $\set{1, \dots, k}$. Distinct subsets give distinct divisors, by uniqueness of factorization.
7. The divisor $d_S$ has exactly $\abs{S}$ distinct prime factors, so $\omega(d_S) = \abs{S}$ (see the number of distinct prime factors). Hence $\mu(d_S) = (-1)^{\abs{S}}$.
8. There are $\binom{k}{j}$ subsets $S$ with $\abs{S} = j$. So $$\sum_{d \mid n} \mu(d) = \sum_{j=0}^{k} \binom{k}{j} (-1)^j.$$
9. By the binomial theorem with $x = -1$ and $y = 1$, this sum equals $(1 - 1)^k = 0$, since $k \ge 1$.
10. So $(\mu * \mathbf{1})(n)$ is $1$ for $n = 1$ and $0$ for $n > 1$. That is, $\mu * \mathbf{1} = \varepsilon$.