Euler Phi Function of Square-Free Integer/Proof 2

Proof
From Euler Phi Function of Integer:
 * $\displaystyle \phi \left({n}\right) = n \prod_{p \mathop \backslash n} \left({1 - \frac 1 p}\right)$

where $p$ ranges over all the primes.

As $n$ is square-free:


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

Hence:
 * $\displaystyle \phi \left({n}\right) = \prod_{p \mathop \backslash n} \left({p - \frac 1 p}\right)$

and so:
 * $\displaystyle \phi \left({n}\right) = \prod_{p \mathop \backslash n} \left({p - 1}\right)$

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

and so:
 * $\displaystyle \prod_{p \mathop \backslash n} \left({p - 1}\right) = \prod_{\substack {p \mathop \backslash n \\ p \mathop > 2} } \left({p - 1}\right)$

Hence the result.