TITLE: Deriving the formula for the Möbius function?
QUESTION [14 upvotes]: There is a rich and beautiful theory in combinatorics which deals with the Möbius functions of posets and arrives at the "Classical" formula of the Möbius function. Let's forget all about it, since I'm hoping to find an elementary solution.
I wish to arrive at the formula for the classical Möbius function $\mu$ (which is defined as $1$ on $1$, $0$ on non square-free numbers, and as $(-1)^k$ for square-free numbers with exactly $k$ prime divisors) from the following formula:
$$\sum_{d|n}\mu(d)=\cases{1, & n=1;\\ 0, & n>1.}$$
So let's assume I'm given this formula and nothing else (especially not a pre-hand knowledge of the definition of $\mu$) - is there a simple way to derive the standard definition, without using deeper theory or "insight"?

REPLY [2 votes]: Here's another solution using Dirichlet series. Consider the ring of sequences $a_n, n \in \mathbb{N}$ under Dirichlet convolution
$$a_n * b_n = \sum_{d | n} a_d b_{n/d}.$$
This ring has identity the sequence $a_1 = 1, a_n = 0, n \ge 2$. We are asked to find the inverse of the sequence $a_n = 1$ in this ring. A basic observation is that the map
$$a_n \mapsto \sum_{n \in \mathbb{N}} \frac{a_n}{n^s}$$
is a ring homomorphism (where the product on power series is the standard product). The sequence $a_n = 1$ is the Riemann zeta function
$$\zeta(s) = \sum_{n \in \mathbb{N}} \frac{1}{n^s}.$$
Another basic observation, the translation to Dirichlet series of the product poset observation in Marc van Leeuwen's answer, is that the zeta function has an Euler product
$$\zeta(s) = \prod_{p \text{ prime}} \left( \frac{1}{1 - p^{-s}} \right)$$
and hence its inverse has Euler product
$$\zeta(s)^{-1} = \sum_{n \in \mathbb{N}} \frac{\mu(n)}{n^s} = \prod_{p \text{ prime}} \left( 1 - \frac{1}{p^s} \right).$$
The conclusion follows by expanding the RHS.