Proof

Let F be a finite field of order q. By previous result there is a prime number p and a positive integer a such that q = p^a.

Suppose m is a natural number dividing q - 1. We show that the number of elements x F^* with x^m = 1 equals m.
It is precisely the number of solutions of X^m - 1 in F. As m | (q - 1), the polynomial X^m - 1 divides X^{q - 1} - 1, whence also X^q - X. By a previous theorem, the latter polynomial decomposes into a product of linear factors in F[X]. But then its divisor X^m - 1 is also a product of m linear factors. Hence, there are m solutions of X^m - 1 in F. In other words, the number of elements x F^* with x^m = 1 equals m.

Finally, we apply the cyclic group characterization theorem to conclude that F^* is cyclic.