There is a quite elementary proof that uses only the fact that a polynomial of degree $n$ has at most $n$ roots in a field.

So, let $F$ be a finite field let $g$ be an element of maximal possible order $n$ in the multiplicative group $F^*=F\setminus\{0\}$ of $F$. Let $H:=\{g^k:0\le k<n\}$ be the cyclic subgroup of $F^*$, generated by the element $g$. If $H=F^*$, then the group $F^*$ is cyclic and we are done. So, assume that $H\ne F^*$. Observe that every element $x\in H$ satisfies the equation $x^n-1=0$, which has at most $n$ roots in the field $F$. Therefore, $H=\{x\in F:x^n=1\}$. Let $p$ be the smallest positive number for which there exists an element $y\in F^*\setminus H$ such that $y^p\in H$. The minimality of $p$ implies that $p$ is a prime number. Let $k\in\{0,\dots,n-1\}$ be the smallest possible number such that $y^p=g^k$ for some element $y\in F^*\setminus H$. We claim that $k<p$. Assuming that $k\ge p$, we can observe that the element $z=yg^{-1}\in F^*\setminus H$ has $z^p=(yg^{-1})^p=g^{k-p}$, which contradicts the minimality of $k$. If $k=0$, then $y^p=g^k=1$ and $y\notin H$ imply that $p$ does not divide $n$. In this case the element $yg\in F^*$ has order $pn>n$, which contradicts the maximality of $n$. This contradiction shows that $k>0$. It follows from $0<k<p$ that the element $g^k$ has order $d>\frac{n}{p}$. Then the element $y$ has order $pd>n$, which contradicts the maximality of $n$. This is a final contradiction showing that the group $F^*=H$ is cyclic.

There is a quite elementary proof that uses only the fact that a polynomial of degree $n$ has at most $n$ roots in a field.

So, let $F$ be a field and $G$ be a finite subgroup of the multiplicative group $F^*:=F\setminus\{0\}$ of $F$. To prove that $G$ is cyclic, choose an element $g\in G$ of maximal possible order $n$ in $G$. Let $H:=\{g^k:0\le k<n\}$ be the cyclic subgroup of $G$, generated by the element $g$. If $H=G$, then the group $G$ is cyclic and we are done. So, assume that $H\ne G$. Observe that every element $x\in H$ satisfies the equation $x^n-1=0$, which has at most $n$ roots in the field $F$. Therefore, $H=\{x\in F:x^n=1\}$. Let $p$ be the smallest positive number for which there exists an element $y\in G\setminus H$ such that $y^p\in H$. The minimality of $p$ implies that $p$ is a prime number. Let $k\in\{0,\dots,n-1\}$ be the smallest possible number such that $y^p=g^k$ for some element $y\in G\setminus H$. We claim that $k<p$. Assuming that $k\ge p$, we can observe that the element $z=yg^{-1}\in G\setminus H$ has $z^p=(yg^{-1})^p=g^{k-p}$, which contradicts the minimality of $k$. If $k=0$, then $y^p=g^k=1$ and $y\notin H$ imply that $p$ does not divide $n$. In this case the element $yg\in G$ has order $pn>n$, which contradicts the maximality of $n$. This contradiction shows that $k>0$. It follows from $0<k<p$ that the element $g^k$ has order $d>\frac{n}{p}$. Then the element $y$ has order $pd>n$, which contradicts the maximality of $n$. This is a final contradiction showing that the group $G=H$ is cyclic.