### The number of elements in vector spaces over finite fields.

Let $V$ be a vector space over a field $F$. If $n=\dim(V)$ is finite, then any choice of a basis $\beta=\{v_1,\ldots,v_n\}$ gives an isomorphism $F^n\cong V$, by the map $$F^n\to V,\ \ (a_1,\ldots,a_n)\mapsto a_1v_1+\ldots +a_n v_n.$$ Indeed, this map is bijective, due to the result that every $v\in V$ is uniquely a linear combination of elements of the basis. It also preserves addition and scalar multiplication. Suppose now that $F$ is a finite field. Just from the fact that the map above is a bijection, we get that $$\# V=\# (F^n)=(\# F)^n.$$

### The number of elements of finite fields.

An important special case: Recall that for every finite field, the unique smallest natural number $p$ such that $$p\cdot 1=1+\ldots+1=0$$ is a prime number. We obtain an inclusion of $\mathbb{Z}_p$ as a subfield of $F$, consisting of elements $0,1,1+1,\ldots$. But then we may regard $F$ as a vector space over the field $\mathbb{Z}_p$. Using the formula above (with $F$ playing the role of $V$ and $\mathbb{Z}_p$ playing the role of $F$), since $\# \mathbb{Z}_p=p$, we find that $$\# F=p^n.$$ So, we proved that the number of elements in a finite field is always some power of a prime number! Note that the statement itself doesn't involve vector spaces and bases, even though the proof does. (It also true that every power of a prime number really occurs as number of elements of a finite field, in fact a unique one, but this fact is harder.)