Let $\mathbb{K}$ be a finite field with cardinality $|\mathbb{K}| = m \in \mathbb{N}$.
Let $V$ be a vector space over $\mathbb{K}$ with finite dimension $n \in \mathbb{N}$ and some base $B = (v_1, \ldots, v_n)$.
Then $V$ is a finite set with cardinality $|V| = m^n$.
Proof.
We know that $V = \text{span}(v_1,\ldots,v_n) = \{\sum_{i = 1}^{n} \lambda_iv_i: \lambda \in \mathbb{K}\}$.
Since there are only finite vectors in the base and only a finite amount of scalars in the field, there are finite-many linear combinations of them, the exact amount being the cardinality of $V$. With elementary combinatorics it follows:
$|V| = |\mathbb{K} \underbrace{\times \ldots \times }_{n \text{ times}} \mathbb{K}| = |\mathbb{K}^n| = |\mathbb{K}|^n = m^n \in \mathbb{N}$.
If $V$ is the null space, meaning $V = \{0\}$, then $V$ is finite-dimensional with dimension $n=0$ and the set $V$ is finite with cardinality $m = 1$, and the claim still holds: $|\{0\}| = 1 =1^0= m^n$.
Is this a correct proof? How could I formalize the idea I articulate in the written paragraph?