number theory - Proof that the nth root of a rational s/t is irrational unless s and t are perfect nth powers

I'm trying to prove that $\sqrt[n]{\frac{s}{t}}$ is irrational unless both s and t are perfect nth powers. I have found plenty of proofs for nth root of an integer but cannot find anything for rationals. Also trying to work up from the proofs I have found is rather difficult.

My lecturer wants me to prove this using uniqueness of prime factorisation but I also have no idea where to start with that.

Any help is greatly appreciated.

Answer

To build off Vasya and Calum's helpful comments:

Let $\frac{s}{t}$ be an irreducible fraction (to address Vasya's comment).

Continuing with the main proof, recall Vasya's comment:

$\sqrt[n]{\frac{s}{t}} = \frac{\sqrt[n]{s}}{\sqrt[n]{t}}$

We will quickly prove this.

Obviously $(\sqrt[n]{\frac{s}{t}})^n = \frac{s}{t}.$

Notice $\frac{s}{t} = \frac{(\sqrt[n]{s})^n}{(\sqrt[n]{t})^n} = (\frac{\sqrt[n]{s}}{\sqrt[n]{t}})^n.$

It follows that $\sqrt[n]{\frac{s}{t}} = \frac{\sqrt[n]{s}}{\sqrt[n]{t}}$ since the $n$th root is either a bijection or always positive.

Without loss of generality, let $s$ have a prime factorization $p_1^{a_1}p_2^{a_2}...p_i^{a_i}.$ Let $\sqrt[n]{s}$ be an integer with prime factorization $q_1^{b_1}q_2^{b_2}...q_j^{b_j}.$

Clearly $p_1^{a_1}p_2^{a_2}...p_i^{a_i} = q_1^{nb_1}q_2^{nb_2}...q_j^{nb_j}.$
We know the prime factorization of $s$ is unique, so we know that each power of $p$ corresponds to some power of $q.$
It's not clear which $p$ corresponds to which $q,$ but it is clear that all the prime factors are $n$th powers.