Intuitively, for every $n \ge 2$ I can find an element $u=a^{1\over{n}}$with $a \in \mathbb{Q}$ and an irreducible polynomial in $\mathbb{Q}[X]$: $u^{n}-a=0$. As $u \notin \mathbb{Q}$, then $[\mathbb{Q}(u)/\mathbb{Q}]=n$ which is a finite extension.
Is this reasoing OK?
Edit: I guess I need to prove that the polynomial is actually irreducible using Eisenstein's criterion.
You are definitely on the right track. You should justify why you are guaranteed an irreducible polynomial, which is simple: let $f(x) = x^n - p$ for some prime $p$. Then $f$ is certainly irreducible by Eisenstein's criterion.
From there, if $\alpha$ is a root of $f$, then $\mathbb{Q}[\alpha]$ is a field extension of dimension $\deg(f) = n$.
I should add that $f(x) = x^n - a$ for any arbitrary $a \in \mathbb{Q}$ is not necessarily irreducible. For example, $f(x) = x^n - 1$ will always have a factor of $(x-1)$.
