Irrationality of $2^{\frac{1}{n}}$ for $n > 2$

Question

What do you guys think about this proof? So, we suppose $$2^{\frac{1}{n}}$$ is rational ( = $\frac{p}{q}$). Therefore, $$2 = \frac{p^n}{q^n} \Rightarrow q^n + q^n = p^n$$ and this contradicts fermat's last theorem. Is this correct proof?

Answer 1

You still need to prove it for $n=2$ though. The proof is right for $n >2$. But this proof falls under the category of mosquito nuking proofs.

Answer 2

Well if $2=p^n/q^n$, then $2\cdot q^n=p^n$. This implies that 2 divides $p$; thus $2^n$ divides $p^n$. This in turn implies that $2^n$ divides $2\cdot q^n$, but since $n>1$, 2 must divide $q$ as well. This is a contradiction.