# Proof that $\sum_{d|m} |\mu(d)|=2^n$, where $n$ is the number of distinct prime divisors of $m$?

Given an integer $m$ such that $n$ is denoting the distinct prime divisors of $m$, is there a proof that the sum over the divisors of m of the absolute value of the Möbius function $\mu(d)$ is equal to

$$\sum_{d|m} \left|\mu(d)\right|=2^n,$$

where the Möbius function is defined as
$$%\small \mu(d) = \left\{ \begin{array}{ll} 1 & \text{if d = 1} \\ (-1)^k & \text{if d is the product of k distinct primes} \\ 0 & \text{if d has one or more repeated prime factors}\,. \end{array} \right.$$

I checked the relation in Maple empirically and it seems to be correct, however I could not come up with a formal proof for this result. Any help is greatly appreciated.

Thanks.

Note that $|\mu(d)|$ is non-zero and equal to one when $d$ is a product of primes, i.e. there is a bijection between those $d$ and the subsets of the set of prime divisors of $m$. There are precisely $2^n$ of these, done.
Denoting the distinct prime factors of $m$ as $p_1, \cdots, p_n$: $\sum_{d|m} \lvert \mu(d)\rvert = \lvert \mu(1)\rvert + \sum_i \lvert \mu(p_i)\rvert + \sum_{1\leq i \leq j \leq n} \lvert \mu(p_i p_j) \rvert + \cdots + \lvert \mu(p_1 p_2\cdots p_n)\rvert$ $= 1 + \binom{n}{1} 1 + \binom{n}{2} 1^2 + \cdots + \binom{n}{n}1^n$