Is $n^\frac{1}{n}$ ever rational?

$n^{\frac{1}{n}}$ cannot be rational for any positive integer $n>1$ (No matter whether $n$ is prime or composite)

This is because the number $n^{\frac{1}{n}}$ is a root of the polynomial $x^n-n$.

The leading coefficient is $1$, hence any rational root woule be an integer. If we denote $m:=n^{\frac{1}{n}}$, we get $m^n=n$. $m$ is clearly positive, so it would have to be a posiive integer, if it were rational.

We would have $m\ne 1$, hence $m\ge 2$, but then $m^n\ge 2^n>n$ for $n>1$, hence we arrive at a contradiction.

$$ 1 < n^{1/n} < 2 \quad \forall n >1 , n\in \Bbb N$$

Also (I think more hint is required as downvotes are too fast) note that $n^{1/m}$ can be rational iff it is an integer.

Consider the polynomial $p$, such that $p(x):= x^n -n $ assume $\sqrt[n]{n} $ as a root.

By the rational root theorem we know that if $p$ has a rational root, it will be the one of the dividers $d_1,\cdots,d_m$ of $n$ (because the coefficient of monomial $x^n$ is $1$). But NONE of then will be a root of $p$. Therefore, the real roots of $p$ are all irracional roots, including $\sqrt[n]n$.