The multiplicative group of a finite field is cyclic

Today our goal is to show (following Dummit and Foote's approach) that the finite subgroup $H$ of the multiplicative group of a field $F$ is cyclic. The title of this post follows as a corollary.

We will use the fundamental theorem of finitely generated abelian groups, found here. This tells us that our subgroup is isomorphic to
$$\mathbb Z/n_1 \mathbb Z \times \cdots \times \mathbb Z / n_k \mathbb Z$$
for some positive integer $k$ and integers $n_k \mid n_{k - 1} \mid \cdots \mid n_1$.

Recall that if $G$ is a cyclic group, say $\mathbb Z / n \mathbb Z$, and $d \mid |G|$, then $G$ contains precisely $d$ elements of order dividing $d$, namely $0, n/d, 2n/d, ..., (d - 1)n / d$.

Now, if $k \geq 2$, there are at least two direct factors of $H$ with $n_k$ elements of order dividing $n_k$, so there are more than $n_k$ elements of the direct product with order dividing $n_k$. But then there are more than $n_k$ roots to the polynomial $x^{n_k} - 1$ in the field $F$, a contradiction. Therefore, $k = 1$, so $H$ is cyclic.