I have proved this for the square root of n, but not for any root of n. To prove this where k=2, I showed that:
If n is a perfect square then $\sqrt n$ is an integer and not irrational.
If $n$ is not a perfect square, assume to the contrary that $\sqrt n$ is rational and can be written as $\frac{a}{b}$.
$\sqrt n$ = $\frac{a}{b}$ where $a,b$ are integers with $GCD(a,b) = 1$
then
$ n = \frac{a^2}{b^2}$ where $b \neq 1$
But $\frac{a^2}{b^2}$ is not an integer so there is a contradiction. So $ \sqrt n \neq\frac{a}{b}$ and $\sqrt n$ must be irrational.
How would I expand this to show that it works for any value of $k$ where $k$ is the $kth$ root of $n$?
In fact, the $k$-th root of $n$ satisfies the equation $x^k - n = 0$.
Therefore it is an algebraic integer. If it is also rational, then it is actually an integer, as $\Bbb Z$ is integrally closed.
The simple proof of that: if $x = \frac a b$ is rational, with $a, b$ coprime integers, then we have $a^k - n b^k = 0$, which leads to $b \mid a^k$.
But $a, b$ are coprime, so it must follow that $b = 1$, and thus $x = a$ is an integer.