Editing Euler's totient function (section)

====Proof of Euler's product formula====

The [[fundamental theorem of arithmetic]] states that if {{math|''n'' > 1}} there is a unique expression <math>n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}, </math> where {{math|''p''<sub>1</sub> < ''p''<sub>2</sub> < ... < ''p''<sub>''r''</sub>}} are [[prime number]]s and each {{math|''k''<sub>''i''</sub> ≥ 1}}. (The case {{math|1=''n'' = 1}} corresponds to the empty product.) Repeatedly using the multiplicative property of {{mvar|φ}} and the formula for {{math|''φ''(''p''<sup>''k''</sup>)}} gives

:<math>\begin{array} {rcl}
\varphi(n)&=& \varphi(p_1^{k_1})\, \varphi(p_2^{k_2}) 
\cdots\varphi(p_r^{k_r})\\[.1em]
&=& p_1^{k_1} \left(1- \frac{1}{p_1} \right) p_2^{k_2} \left(1- \frac{1}{p_2} \right) \cdots p_r^{k_r}\left(1- \frac{1}{p_r} \right)\\[.1em]
&=& p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r} \left(1- \frac{1}{p_1} \right) \left(1- \frac{1}{p_2} \right) \cdots \left(1- \frac{1}{p_r} \right)\\[.1em]
&=&n \left(1- \frac{1}{p_1} \right)\left(1- \frac{1}{p_2} \right) \cdots\left(1- \frac{1}{p_r} \right).
\end{array}</math>

This gives both versions of Euler's product formula.

An alternative proof that does not require the multiplicative property instead uses the [[inclusion-exclusion principle]] applied to the set <math>\{1,2,\ldots,n\}</math>, excluding the sets of integers divisible by the prime divisors.