How to prove that $GF(p^n)$ contains an element of degree n? I am reading Gallian's Algebra book, and I am lapsing on Corollary 2 to Theorem 22.2. The statement goes: "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 $n$." I totally understand the proof in which it says $[GF(p)(a):GF(p)]=[GF(p^n):GF(p)]=n$. But why can we use this argument to show that $a$ is algebraic over $GF(p)$? I only know that we can deduce $[F(a):F]=n$ if $a$ is algebraic over $F$ with minimal polynomial of order $n$, but not the other way around...

Every element of $GF(p^n)$ is algebraic over $GF(p)$. Indeed, if $x\in GF(p^n)$ is nonzero, then $x^m=1$ for some $m$, and so $x$ is a root of the polynomial $x^m-1$. (I'll leave the case that $x=0$ to you.)