Multiplicative group of a finite field is cyclic

Statement

Suppose $F$ is a finite field. Let $F^*$ denote the multiplicative group of $F$ . Then, $F^*$ is a cyclic group.

Note in particular that any finite field has order a prime power, say $q = p^r$ for a prime number $p$ and positive integer $r$ . The multiplicative group is therefore a cyclic group of order $q - 1 = p^r - 1$ .
Since the multiplicative group is cyclic, there are $\varphi(q - 1)$ many choices for the generator of the group. Here, $\varphi$ is the Euler totient function.
Multiplicative group of a finite prime field is cyclic (see also classification of natural numbers for which the multiplicative group is cyclic)

For $r > 1$ , any generator of the multiplicative group is also a primitive element for the field of $q$ elements as an extension of its prime subfield (of $p$ elements). (A primitive element for a field extension is an element that, when adjoined to the smaller field, generates the larger field). However, not every primitive element is a generator of the multiplicative group. In fact, the number of generators of the multiplicative group could be substantially smaller than the number of primitive elements. For instance, consider the case $p = 5, r = 2$ . The multiplicative group has $\varphi(24) = 8$ generators, whereas the field has $p^2 - p = 25 - 5 = 20$ primitive elements.
Every finite division ring is a field