I think of it like this:
We let $K$ be a vector field over $F$. Then, every $\alpha\in K$ be represented uniquely as $$ \alpha = l_1k_1 + \ldots l_nk_n $$ where $k_1,\ldots, k_n$ is a fixed basis of $K$, and $l_1,\ldots, l_n\in L$.
Thus, for each $k_i$, we have a choice of $q$ scalars. Thus, we have $q^n$ different linear combinations. Since each vector has a unique representation, and every linear combination must be contained in our vector space, $K$ has $q^n$ elements.
Is there anything wrong with this proof?