Let $a$ be a generator of the group of nonzero elements of $GF(p^n)$ under multiplication. Then $a$ is algebraic over $GF( p)$ of degree $n$.
The proof is given as observe that $[GF( p)(a):GF( p)] = [GF( p^n):GF( p)] = n.$
My question is, how is it that $[GF( p)(a):GF( p)] = [GF( p^n):GF( p)]$? I see that $GF(p)(a)$ is a vector spaces over $GF(p)$ that contains $1, a,a^2, ..., a^{p-1}$, but I can't see how they're equal.
Consider how $F=GF(p^n)^\times$ is a group, so by LaGrange's theorem, all elements of $F$ satisfy $x^{p^n}-x=0$. Since your element generates $F$, its multiplicative order is exactly $p^{n}-1$, i.e. it does not satisfy $x^{k}-1=0$ for any $k<p^n-1$. But then if $\alpha\in E\subset F$ is in a proper subfield where here $E=GF(p^j)$ and $j<n$ then $\alpha^{p^j-1}-1=0$ a contradiction. So it must be that $\alpha$ generates the field $F$. As you have noted the degree is $n$ for that extension, so $\alpha$ must have minimal polynomial of degree $n$ as well.