Field Generated by Finitely Many Algebraic Elements is Algebraic

Proof

First note that $\mathbb{F}({\alpha }_1)/\mathbb{F}$ is finite due to the power basis.

Then, if $\mathbb{F}({\alpha }_1,\dots ,{\alpha }_k)$ is finite so is $\mathbb{F}({\alpha }_1,\dots ,{\alpha }_k)({\alpha }_{k+1})=\mathbb{F}({\alpha }_1,\dots ,{\alpha }_{k+1})$ since ${\alpha }_{k+1}$ is algebraic over $\mathbb{F}$ which is a subfield of $\mathbb{F}({\alpha }_1,\dots ,{\alpha }_k)$.

Thus the result follows by induction on $k$.