Is $\mathbb{F}$ an algebraic extension of $k$ if $\dim_k V < |k|$?

Question

Let $k \subset \mathbb{F}$. $k, \mathbb{F}$ are fields. Vector space over $\mathbb{F}$ isn't trivial ($V_\mathbb{F} \neq 0$). $\dim_k V < |k|$. Is $\mathbb{F}$ an algebraic extension of $k$?

So I should prove that $\mathbb{F}$ got no transcendent elements over $k$. I don't know where to start.

Answer 1

If $F$ had a transcendental element $x$ then it has these $k$-linearly independent elements: $1/(x-a)$, $a \in k$, so ${\rm dim}_k(V) \geq {\rm dim}_k(F) \geq |k|$, contradicting the assumptions. Thus $F$ must be algebraic over $k$.