Let z = aⁿ -1 Clearly z = (a-1)(a^n-1 + a^n-2 + ..+ a + 1) when a is not 2.

Therefore z is always composite if a > 2

If a = 2 and n is even then clearly n = 2v so that z = (2^v +1)(2^v -1)

If a = 2 and n is uneven , but composite then let n = bc

so that z = 2^bc -1 = f^c - 1 where f = 2^b

It is clear that f is not 2 and that f^c - 1 is therefore composite.

Therefore if 2ⁿ - 1 is prime then n is not even or odd composite because then 2ⁿ - 1 would be composite.

It follows then that n must be prime if 2ⁿ - 1 is prime.