Showing that any field extension of a finite field is simple 
We know that the multiplicative subgroup $F^\times$ of a finite field $F$ is cyclic. Use this to show that any field extension of a finite field is simple. 

Any clues?
 A: As Bernard points out, your statement is false the way you've written it. For example, the fields $\overline{\mathbb{F}_p}$ and $\mathbb{F}_p(x,y)$ are not simple extensions of $\mathbb{F}_p$.
However, what you probably meant to write is that if $F$ is a finite field, then any field extension $K/F$ of finite degree is simple. This is true. Since the degree $[K:F]$ being finite implies that $K$ is itself a finite field, we know that $K^\times$ is cyclic. Let $u$ be a generator of $K^\times$. Then $K=F(u)$ because $F(u)$ has at least $|K|-1$ elements, since the multiplicative order of $u$ is $|K^\times|$. But since $F(u)$ is an $F$-vector space, it must have at least $|K|$ elements, and since $F(u)$ is a subfield of $K$, it must be equal to $K$. Thus the extension $K/F$ is simple.
A: It is false: if it were, any algebraic extension of a finite field $F$ would be a finite extension of $F$, hence its algebraic closure would be a finite field. It's easy to show an algebraically closed field can't be finite.
