Euler Phi Function of Square-Free Integer/Proof 2

Proof
From Euler Phi Function of Integer:
 * $\displaystyle \map \phi n = n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}$

where $p divides n$ denotes the primes which divide $n$.

As $n$ is square-free:


 * $\displaystyle n = \prod_{p \mathop \divides n} p$

Hence:
 * $\displaystyle \map \phi n = \prod_{p \mathop \divides n} \paren {p - \frac 1 p}$

and so:
 * $\displaystyle \map \phi n = \prod_{p \mathop \divides n} \paren {p - 1}$

When $p = 2$ we have that:
 * $p - 1 = 1$

and so:
 * $\displaystyle \prod_{p \mathop \divides n} \paren {p - 1} = \prod_{\substack {p \mathop \divides n \\ p \mathop > 2} } \paren {p - 1}$

Hence the result.