Irrationality of ⁿ√2

Assume $\sqrt[n]{2}=p/q$ where $p$ and $q$ are positive integers such that $\gcd(p,q)=1$. Then $p^n=2q^n$ and $P$ is even; write $p = 2p'$ where $p$ is a positive integer. That implies $2 q^n = p^n = 2^n p'^n$, hence $q^n = 2^{n-1} p'^n$, and thus $q$ is even (because $n \geq 2$). This is impossible since $\gcd(p,q)=1$.