Let $K$ be a field and let $x\in K$. Let $p$ be prime and assume that $x$ is not a $p$-th power. Now let $y\in\overline{K}$ be an element such that $y^p=x$ and define $d:=[K(y):K]$. Here I use the notation $\overline{K}$ as some algebraic closure of $K$. I've already shown that $2\leq d\leq p$, and I don't really see where it could help more to use $[K(y):K]$ instead of just $d$ (which is why I'll just use $d$). Now suppose that we have the equality $x^d=x^p$. Do we then necessarily have $d=p$?
From here I don't really know what to do anymore. The only cases I can think of are those where $x$ is either $0$ or $1$, but since these are $p$-th powers $(0^p=0$ and $1^p=1)$ it gives me the feeling the statement might be true. I can't think of any proof or counterexample, can anyone help me with the question? Thanks in advance!
