I am asked to prove that $L=\bigcup_{n=1}^\infty\mathbb{Q}(\sqrt[n]2)$ is an algebraic field extension over $\mathbb{Q}$. So far I have:
Let $\beta\in L$, then by definition of union there exists a $N\in\mathbb{Z}_{\geq1}$ such that $\beta\in\mathbb{Q}(\sqrt[N]{2})$. $\sqrt[N]{2}$ is algebraic in $\mathbb{Q}$ (take $f=X^N-2$), so $\beta$ is algebraic as well. Since $\beta$ was chosen arbitrarily, every element of $L$ is algebraic over $\mathbb{Q}$. This makes $L$ an algebraic field extension.
All examples I've seen so far are finite extensions, and $L$ is obviously not finite. I'm wondering if this proof suffices, or if I'm missing something important.