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Homework Statement
Let phi denote Euler's totient function, ie for k in N, phi(k) is the number of natural numbers less than k which are relatively prime to k.
Prove that for every n in N we have n| phi((2^n) -1).
(Hint: Compute o[2] in U_(2^n)-1, the group of units in Z mod (2^n)-1)
The Attempt at a Solution
I understand that in the hint, o[2]=n...but I don't know how to relate it to the problem.
I also found that the value of phi(k) for any k is an even integer.