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:
 * $\displaystyle n = \prod_{1 \mathop \le i \mathop \le r} p_i = p_1 p_2 \cdots p_r$

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:
 * $\displaystyle \phi \left({p_i}\right) = \left({p_i - 1}\right)$

Thus:
 * $\displaystyle \phi \left({n}\right) = \prod_{1 \mathop \le i \mathop \le r} \left({p_i - 1}\right)$

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

where $p$ ranges over all prime numbers

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.