Irrationality of $\sqrt[n]2$ [duplicate]

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

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.

Apply Eisenstein's criterion with $p = 2$ to the polynomial $f(x) = x^n - 2$. This shows that $f$ is irreducible over $\mathbb{Q}$, so in particular it has no roots in $\mathbb{Q}$.

FLT is quite a cannon to settle the proof. One can also follow the "classical" approach: assuming WLOG $\gcd(p,q)=1$, if $p^n=2q^n$, then $2 \mid p^n$, implying $2 \mid p$, say $p=2r$. But then $2^{n-1}r^n=q^n$, and since $n \gt 2$, we see that $2 \mid q^n$. It follows that $2 \mid q$, contracting the fact that $p$ and $q$ are relatively prime.