How can we find $[GF(p^n):GF(p)]=n$?

Question

I was searching why $[GF(p^n):GF(p)]=n$. It is not very logical, isn't it ? I know that $$GF(p^n)=\{x\in GF(p)^{alg}\mid x^{p^n}=x \}$$ is a field with $p^n$ element since it split $X^{p^n}-X$ which is separable. But how can we get an irreducible polynomial in $GF(p)$ of degree $n$ and that split over $GF(p^n)$ ?

Answer 1

There is no simple formula for an irreducible polynomial of degree $n$. The fact that the degree of the extension is $n$ is immediate for reasons of cardinality.

You said correctly the cardinality of $GF(p^n)$ is $p^n$. And the cardinality of a vectorspace of dimension $d$ over $GF(p)$ is $p^d$. Thus the dimension of $GF(p^n)$ as a $GF(p)$ vectorspace must be $n$.