WTS: For any positive real number $n$ such that $n$ is not a perfect $k$-th power, $\sqrt[k]{n}$ is irrational.
Case 1: $n$ is rational
Let $n=\frac{p}{q}$ for two co-prime, positive integers $p$ and $q$ such that $n\neq a^k$ for some integer $k$ and rational $a$. Let $d_1$, $d_2$, $\cdots$, $d_{m-1}$, and $d_m$ be the positive divisors of $p$. Similarly, let $f_1$, $f_2$, $\cdots$, $f_{h-1}$, and $f_h$ be the positive divisors of $q$. Define a function $g(x)$ such that $$g(x)=qx^k -p.$$ Clearly, the only positive root of $g(x)$ is $\sqrt[k]{\frac{p}{q}}$. Additionally, by the Rational Root Theorem, the only possible positive, real, rational roots of $g(x)$ are $\frac{d_i}{f_j}$ for integers $1\leq i \leq m$ and $1\leq j \leq h$. Now, we have a different question to ask: does the equation $$\frac{d_i}{f_j}=\sqrt[k]{\frac{p}{q}}$$ have any solutions? Let's first assume that it does. Raising both sides to the $k$-th power, $$\frac{p}{q}=n=\left(\frac{d_i}{f_j}\right)^k.$$ Because $d_i$ and $f_j$ are integers, $\frac{d_i}{f_j}$ is rational. However, this contradicts the fact that $n\neq a^k$ for a rational $a$. Hence, $\sqrt[k]{n}$ is irrational. $\square$
Case 2: $n$ is irrational
Let $n$ be a real, irrational number. Assume $\sqrt[k]{n}$ is rational (i.e. $\sqrt[k]{n}$ can be expressed as the ratio of two co-prime integers, $r$ and $s$). $$\sqrt[k]{n}=\frac{r}{s}$$ Raising both sides to the $k$-th power, $$n=\left( \frac{r}{s} \right)^k=\frac{r^k}{s^k}.$$ However, because $r$, $s$, and $k$ are all integers, $r^k$ and $s^k$ are both integers. Therefore, $n$ must be rational, a contradiction. Hence, $\sqrt[k]{n}$ is irrational. $\blacksquare$