Prove that $[GF(p^n):GF(p)] = n$
The proof is:
Since $\Bbb Z_p \oplus \Bbb Z_p \oplus \dots \Bbb Z_p$ is a vector space over $\Bbb Z_p$ with $\{(1, 0, . . . , 0), (0, 1, 0, . . . , 0), > . . . , (0, 0, . . . , 1)\}$ as a basis, we have $[GF(p^n):GF(p)] = > n$.
My question is: Why does it matter that $\Bbb Z_p \oplus \Bbb Z_p \oplus \dots \Bbb Z_p$ is a vector space over $\Bbb Z_p$ with $\{(1, 0, . . . , 0), (0, 1, 0, . . . , 0), . . . , (0, 0, . . . , 1)\}$ as a basis?
Because $GF(p^n)$ is only isomorphic to $\Bbb Z_p \oplus \Bbb Z_p \oplus \dots \Bbb Z_p$ when $GF(p^n)$ is a group under addition. So technically $[\Bbb Z_p \oplus \Bbb Z_p \oplus \dots \Bbb Z_p : \Bbb Z_p] = n$ should be a correct statement because this satisfies the vector space condition.
Since $GF(p^{n})$ is finite,$[GF(p^n):GF(p)]$ is finite.Suppose$[GF(p^n):GF(p)]=m\neq n$,and then each element of $GF(p^n)$ can be uniquely expressed as $a_{1}v_{1}+a_{2}v_{2}+\cdots +a_{m}v_{m}$,and for each $a_{1},a_{2},\cdots,a_{m}\in GF(p)$,therefore,it is the consequence that $|GF(p^n)|=p^m\neq p^n$,reached a contradiction,thus $[GF(p^n):GF(p)]=n$.Moreover,if field $F$,$H$ are finite,and $[H:F]=n$,then $|H|=|F|^n$.