abstract algebra - Irrationality of $sqrt[n]2$

I know how to prove the result for $n=2$ by contradiction, but does anyone know a proof for general integers $n$ ?

Thank you for your answers.

Marcus

Answer

Suppose that $\sqrt[n]2$ is rational. Then, for some $p,q\in\mathbb Q$,

$$\sqrt[n]2=\frac{p}{q}\implies 2=\frac{p^n}{q^n}\implies p^n=2q^n=q^n+q^n.$$

Contradiction with Fermat last theorem.