A question about finite field extension of a finite field Let $K$ be a finite extension field of a finite field $F$. Show that there is an element $a\in K$ s.t. $K = F(a)$.

My attempt: 

$K$ is a finite field and $char(K) = char(F) := p$. I know that for a given prime $p$ and $n\in N$, there exists a unique finite field upto isomorphism of order $p^n$. 

1) But how can I show that $K \simeq GF(p^n)$ where $p = char(K)$? 

Once I have that, $K\setminus \{0\}$ is a cyclic group under multiplication and thus $\exists a\in K$, s.t. $K\setminus \{0\} = &lt;a&gt;\subset F(a)$. Thus $K\subset F(a)$. 

2) Also, how can I show that $F(a)\subset K$ so that $F(a) = K$? $K$ is an finite field extension of $F$ but I can't understand why that would imply $F(a)\subset K$??
Let $a$ be a generator of the multiplicative group $K^\times$. The elements $a,a^2,...,a^{|K|-1}$ are all different elements in $F(a)$ and hence $|F(a)|\geq |K|-1$. On the other hand $F(a)\subseteq K$ and hence $|F(a)|\leq |K|$. Since the order of $F(a)$ must be a power of $p$ (and $|K|-1$ is not a power of $p$) we conclude that $|F(a)|=|K|$. Since $F(a)\subseteq K$ and both fields are finite this implies $F(a)=K$.
Since the finite extension of a finite field is also a finite field, so there are at most finitely many middle fields $E_i$ such that $F\subset E_i\subset E$. so $E/F$ is a simplicial extension.