How to prove that order of a field is a power of its characteristic?

Question

How to prove that order of a field $\mathbb{GF}(q)$ is a power of its characteristic $\lambda$?

What I can say is that every field has a subfield of order $\lambda$ if it helps

Answer 1

Hint 1: call $K$ the subfield of order $\lambda$. Then $\mathbb{GF}(q)$ becomes a vector space over $K$.
Hint 2: if $x_1,\dots,x_m$ is a basis, every element can be written uniquely as $\sum_{i=1}^m\alpha_i x_i$ with $\alpha_i\in K$. How many $m$-uples $(\alpha_1,\dots,\alpha_m)$ are there?

Answer 2

${\mathbb{GF}}(q)$ is a vectorspace over ${\mathbb{GF}}(\lambda)$ [follows trivially from the definition of vectorspace]. Therefore it is, as a vectorspace, isomorphic to ${\mathbb{GF}}(\lambda)^d$ for some $d$ [by picking a basis]. So it has $\lambda^d$ elements.