I've come across some trouble in a proof and feel like I need a little push to reach the contradiction I'm aiming for!
My goal is to show equality of the fields in the tower $K\subseteq K(a^{1/p})\subseteq K(a^{1/p^2})$, where $p$ is a prime not equal to the characteristic of $K$ and $K(a^{1/p})=K(a^{1/p^2})$ by assumption. Additionally, we have established that $K(a^{1/p})$ is the splitting field of $x^p-a$ and $K(a^{1/p^2})$ the splitting field for $x^{p^2}-a$. [MAJOR EDIT: $K$ contains a primitive $p$th root of unity and if $p=2$, also a primitive fourth root of unity.]
Suppose for sake of contradiction that $a\in K$ is not a $p$th power. Then in the tower above, we have degrees $[K(a^{1/p}):K]=p$ and $[K(a^{1/p^2}):K(a^{1/p})]=1$. Then $[K(a^{1/p^2}):K]=p$, and the minimal polynomial of $a^{1/p^2}$ over $K$ (call it $m(x)$) has degree $p$.
From this point I'm stuck. I know $x^{p^2}-a$ is a polynomial over $K$ that has $a^{1/p^2}$ as a root and thus $m(x)$ must divide this polynomial. Additionally, we have the polynomial $x^p-a^{1/p}$, which is a degree $p$ polynomial with $a^{1/p^2}$ as a root, yet it is over $K(a^{1/p})$ and not $K$. It seems like the answer is staring me in the face, but I haven't found it. If you have any hints to get me there, I would very much appreciate them!