Proof that the multiplicative group $\boldsymbol{F^\times}$ of a finite field $\boldsymbol F$ is cyclic:
Remember that in a group, the exponent
is the least positive integer $n$, if it exists, such that $x^n = 1$ (the unit element of the group). This of course implies all elements in the group have finite order. It is equivalent to say all elements have finite order and the orders have a least common multiple (which is not necessarily true if the group is infinite).
In a finite group, with exponent $e$, there exists an element $\alpha$ of order $e$. So in a finite field, the non-zero elements satisfy the equation $x^e-1=0$ If $F$ is a field with order $q$ ($=p^n$), the equation $x^e-1=0$ has at most $e$ solutions, whence $e\ge q-1$ and finally $e=q-1$, so $F^\times$ is generated by $\alpha$.
Note:
One can actually prove more:
In a field $F$ (finite or infinite), any finite subgroup of the multiplicative group $F^\times$ is cyclic.