p-th root does not become a p-th power when adjoined? Suppose $k$ is a number field of characteristic zero, and $u$ is a unit of infinite order, which is not a $p$-th power in $k$.  Show $\sqrt[p]{u}$ is not a $p$-th power in $k(\sqrt[p]{u})$.  (You can use $\zeta_p\in k$ if it helps).
This seemingly simple question, but I have no clue where to start.  I would prefer a non-computational proof, if possible.  (Note: this question has been edited to reflect the case not covered by the answer below.)
 A: Answer to new question:
The norm of $\sqrt[p]{u}$ with respect to the field extension $k\subset k(\sqrt[p]{u})$ is $(-1)^{p-1}u$.  Assume that the characteristic of $k$ is not $2$.
If $\sqrt[p]{u}$ is a $p$-th power in $k(\sqrt[p]{u})$, then its norm must be a $p$-th power in $k$.  Since $u$ is not a $p$-th power, this can only happen if $p=2$ and $-u = s^2$ is a square in $k$.
In this case, a necessary and sufficient condition for $\sqrt{u}$ to be a square in $k(\sqrt{u})$ is that $\pm 2s$ also be a square in $k$, i.e. that $-4u$ be a fourth power.  We can see this by direct computation of $\sqrt{u} = (a+b\sqrt{u})^2 $, leading to the equations $a = \pm sb$ and $\pm 2sb^2 = 1$.
For a concrete counterexample, take $u=-4$, so that $\sqrt{u} = (1+\frac{1}{2}\sqrt{u})^2$.

Answer to old question:
If $k$ has characteristic $p$, then any $p$-th power in $k(\sqrt[p]{u})$ can be shown, directly, to lie in $k$.  Since $\sqrt[p]{u}\notin k$, it is not a $p$-th power.
If $k$ does not have characteristic $p$, this is false in general, e.g. $-1$ is not a square in $\mathbb{R}$, but $i$ is a square in $\mathbb{C}$.
