The finite field extension Let field $K$ embedded into the finite field $M$. Prove that $M = K(\theta)$ for some $\theta \in M$.
I have tried 2 ways but got stuck at both.
1) Let $|K| = p^s$ and $|M| = p^{st}$ for prime $p$ and $s, t \in \mathbb N$. And if we find the irreducible polynomial $f(x) \in K[x]$ where $deg f(x) = t$, and $f(x)$ has at least one root in $M$ we will solve the problem. But I can not prove there is always such polynomial.
2) The second idea was to represent $K$ and $M$ as $L(\theta)$ where $L = \{n \cdot 1: n = 0,1,\ldots,p-1\}$, where $p$ is characteristic of fields. And if $K = L(\theta_1)$ and $M = L(\theta_2)$ we can say that $M = K(\theta_2)$. But I do not know how to prove that we always can represent a finite field as $L(\theta)$ where $L = \{n \cdot 1: n = 0,1,\ldots,p-1\}$, and moreover I do not know is it truth or not.
Thank you for your help!

One easy approach is to appeal to the following theorem:

If $F$ is any field, and $G$ is a finite subgroup of $F^\times$, then $G$ is cyclic.

Since $M$ is finite, we can take $G=M^\times$, so there is some $\theta\in M$ such that every non-zero element of $M$ is equal to $\theta^n$ for some $n$. Clearly, this $\theta$ will have the property that $M=K(\theta)$.
