One knows the easy
Fact. $$\lim_{n\to\infty}n^{\frac{1}{n}}=1.$$
My proof is defining $l=\lim_{n\to\infty}n^{1/n}$ then take logarithms to show that $\log l=\lim_{n\to\infty}\frac{\log n}{n}=0$, and thus exponentiation yields previous Fact.
Let the radical of an integer defined as you see in previous Wikipedia by $\operatorname{rad}(1)=1$, and when $n>1$ then $$\operatorname{rad}(n)=\prod_{p\mid n}p,$$ that is the product of all distinct primes dividing $n$ (examples are $\operatorname{rad}(29)=29$ and $\operatorname{rad}(28)=\operatorname{rad}(4\cdot 7)=2\cdot 7 $ since this arithmetic function is multiplicative).
Inspired in different calculations and experiments, I've considered the following
Question. Can you provide us a characterization of a sequence of positive integers with $$n^{\frac{1}{\operatorname{rad}(n)}}$$ unbounded? It is: is it possible a (an infinite) sequence $n_k$ of positive integers such that $$n_k^{\frac{1}{\operatorname{rad}(n_k)}}$$ tends to infinite?
Many thanks. I was interested in this question after experiments also with different arithmetic functions, I say divisors functions.
An example is $n_k=2^k$: $n_k^{\frac1{rad(n_k)}}=\sqrt2^k\to\infty$ as $k\to\infty$.
In general, a necessary condition is that the maximal exponent of a prime in the factorisation of $n_k$ is unbounded, because if $n_k=p_1^{a_1}\cdots$ with $a_1$ maximal (say), $n_k^{\frac1{rad(n_k)}}\leq rad(n_k)^{\frac{a_1}{rad(n_k)}}\leq\sqrt2^{a_1}$.
However this is not sufficient, e.g. $n_k=2^k\cdot p_2\cdots p_{k+1}$ has $n_k^{\frac1{rad(n_k)}}$ bounded.