If $C$ is a vector space over a finite field $\mathbb{F_q}$ where q is prime, and the number of different vectors in the vector space is $q^n, n\in \mathbb{N}$, then $dim(C)=n$ ?
It seems to me very simple, since the if $dim(C)>n$ then $|C|>q^n$ which contradicts that $|C|=q^n$, and $dim(C)<n$ then $|C|<q^n$ which again contradicts that $|C|=q^n$. The question is whether it's the right way, and if I need to prove something in addition to what I said?
You could reason as follows: any finite dimensional vector space $V$ of dimension $r$ over a field $K$ is necessarily isomorphic to $K^r$, and thus when $K$ is finite, we will have $|V|=|K^r|=|K|^r$.
Therefore if $|C|=q^n$, and $|K|=|\mathbb F_q|=q$, then the dimension of $C$ must be $n$.
Of course, your argument also works (and is probably equivalent), but you need to explain why $\operatorname{dim}C>n$ implies $|C|>n$ and similarly for the other case. It's not so clear how the logic flows, as it currently stands.