Euler Phi Function of Square-Free Integer/Proof 1

Proof
We have that the Euler Phi Function is Multiplicative.

Let the prime decomposition of $n$ be:
 * $\ds n = \prod_{1 \mathop \le i \mathop \le r} p_i = p_1 p_2 \cdots p_r$

From the definition of prime number, each of the prime factors of $n$ is coprime to all other divisors of $n$.

From Euler Phi Function of Prime, we have:
 * $\map \phi {p_i} = \paren {p_i - 1}$

Thus:
 * $\map \phi n = \ds \prod_{1 \mathop \le i \mathop \le r} \paren {p_i - 1}$

or:
 * $\map \phi n = \ds \prod_{p \mathop \divides n} \paren {p - 1}$

where $p$ ranges over all prime numbers

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

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

Hence the result.