If prime p doesn't divide the class number, then if I is an ideal of $O_K$, and $I ^{p}$ is principal, then I is principal

Question

If a prime p doesn't divide the class number of a number field K, then if I is a non-zero ideal of $O_K$, and $I ^{p}$ is principal, then I is principal.

Answer 1

To expand slightly on my comment, although I'm only familiar with the quadratic case of number fields, I'm pretty sure the class number is the size of the ideal class group. And if $a^p=e$ for $a,e$ in some group, $e$ the identity. Then the order of $a$ divides $p$, but it also always divides the order of the group. Then the order of $a$ divides the gcd of $p$ and the size of the group, which in this case is 1. In this case that implies that the class of $I$ is the identity, so $I$ is principal.