In fact this is a consequence of a more general theorem.

A finite subgroup of the multiplicative group of a field is cyclic. In particular, if F is a finite field, then the multiplicative group of nonzero elements of F is a cyclic group.

By The Fundamental Theorem, the finite subgroup can be written as a direct product of cyclic groups

Where .

In general, if G is a cyclic group and , then G contains precisely d elements of order dividing d. Since divides the order of each of the cyclic groups in the direct product, it follows that each direct factor contains elements of order dividing . If k were greater than 1, there would therefore be a total of more than such elements. But then there would be more than roots of the polynomial in the field F, which contradicts the fact that a polynomial of degree n can have at most n roots. Thus k=1 and the group is cyclic.

Notice is a finite field, so your question is a trivial corollary of the above theorem.