I asked for minimal hints in this question. Now I've come up with a proof. Could you please verify if it is fine or contains logical mistakes?
Let $F$ be a finite field and $F^\times = F \setminus \{0\}$. Then the multiplicative group $F^\times$ is cyclic.
My attempt:
Let $n = |F^\times|$ and $l = \operatorname{lcm}\{\operatorname{order}(x) \mid x \in F^\times\}$. We need the following lemma:
Let $G$ be an abelian group with elements $x, y$ of orders $m$ and $n$ respectively. There exists $z \in G$ of order $\operatorname{lcm} (m,n)$.
Applying this lemma repeatedly, we have that there is $z \in F^\times$ such that $\operatorname{order}(z) = l$.
We now consider the polynomial $X^l -1 \in F[X]$. Because the equation $X^l -1 = 0$ has exactly $n$ different roots, we have $l \ge n$. By Lagrange's theorem, $n$ is divisible by $\operatorname{order}(x)$ for all $x \in F^\times$, so $l \le n$. As a result, $l = n$. In conclusion $F^\times$ is a cyclic group generated by $z$.
Best Answer
Looks fine to me, this is what I had in mind when I posted the hints!
Depending on what level you are seeing this, you might want to justify the smaller details. For example, you use the fact that in a field, the number of roots is at most the degree of the polynomial.
The above seems to really be the crux of the problem and that's where you use the fact that $F$ is a field.