Editing Euler's totient function (section)

==== Value of phi for a prime power argument ====

If {{mvar|p}} is prime and {{math|''k'' ≥ 1}}, then

:<math>\varphi \left(p^k\right) = p^k-p^{k-1} = p^{k-1}(p-1) = p^k \left( 1 - \tfrac{1}{p} \right).</math>

''Proof'': Since {{mvar|p}} is a prime number, the only possible values of {{math|gcd(''p''<sup>''k''</sup>, ''m'')}} are {{math|1, ''p'', ''p''<sup>2</sup>, ..., ''p''<sup>''k''</sup>}}, and the only way to have {{math|gcd(''p''<sup>''k''</sup>, ''m'') > 1}} is if {{mvar|m}} is a multiple of {{mvar|p}}, that is, {{math|1=''m'' ∈ {{mset|1=''p'', 2''p'', 3''p'', ..., ''p''<sup>''k'' − 1</sup>''p'' = ''p''<sup>''k''</sup>}}}}, and there are {{math|''p''<sup>''k'' − 1</sup>}} such multiples not greater than {{math|''p''<sup>''k''</sup>}}. Therefore, the other {{math|''p''<sup>''k''</sup> − ''p''<sup>''k'' − 1</sup>}} numbers are all relatively prime to {{math|''p''<sup>''k''</sup>}}.