The finite field GF(2^k) has order (number of elements) 2^k, but its multiplicative group has the (2^k)-1 nonzero elements, and thus order (2^k)-1, which square

The finite field GF(2^k) has order (number of elements) 2^k, but its multiplicative group has the (2^k)-1 nonzero elements, and thus order (2^k)-1, which squares [pun intended] with the fact that x^(2^k-1) = 1.