Proof by contradiction, Assume $2^{1/n}$ is rational so:
$$2^{1/n} = \frac ab $$ where a,b have no common factors.
$$2 = \frac{a^n}{b^n}$$
$2$ divides LHS, therefore $2$ divides RHS so $2$ divides $a^n$ or $2$ divides $b^n$ which implies $2$ divides $a$ or $2$ divides $b$.
Stuck on what to do next.
Factor $a$ and $b$ into products of primes. We have the identity $2b^n = a^n$; compare the exponents of the primes on both sides of the equation (and look in particular at the exponent of 2).
Let $n \geq 2$; let there be some rational $a,b > 0$ such that $2^{1/n} = a/b$ with $\gcd (a,b) = 1$. Note that $2^{1/n} = a/b$ iff $2b^{n} = a^{n}$; so $2$ divides $a^{n}$, and hence $a$ must be even. Then $n \geq 2$ implies that $4$ divides $a^{n}$; so $2$ divides $b^{n}$ and hence $b$ is even too. But this is a contradiction.
Write:
$a^n= 2 b^n$ simply by working out the fraction.
This implies $a^n$ is even, and thus $a$ must be even (make sure you understand why).
Now we write $a = 2k$ and our relation now reads:
$(2k)^n = 2^n k^n = 2 b^n$ thus for $n>1$ we have:
$b^n = 2^{n-1} a^n$.
Now, this directly forces $b^n$ to be even and thus $b$ must be even.
Concluding:
$$2^{1/n} = \frac{a}{b}$$ is a fraction of even numbers and thus not reduced!
