The irreducibility of $x^p-a$ implies that of $x^{p^2}-a$ 
Morandi's Field and Galois Theorey, exercise 10.5c
Let $p$ be a prime, and suppose that either $F$ contains a primitive $p$th root of unity for $p$ odd, or that $F$ contains a primitive fourth root for $p=2$. If there is an $a\in F$ with $x^p-a$ irreducible over $F$, then $x^{p^2}-a$ is irreducible over $F$. (Hint: Use a norm argument)

My efforts: suppose $\alpha^{p^2}-a=0$, and let $K=F(\alpha),\beta=\alpha^p,L=F(\beta)$ Since $x^p-a$ is irreducible, we have $[L:F]=p$. If $[K:L]=p$, we have $[K:F]=p^2$ and therefore $x^{p^2}-a$ is irreducible, so we only need to show that $[K:L]=p$, which is equivalent to the irreducibility of $x^p-\beta$. Suppose not, since a $p$th primitive root is contained in $L$, we have $x^p-\beta$ splits in $L$, which means that $\alpha\in L$, therefore $\alpha$ is an $F$-polynomial of $\beta$.
I don't know how to proceed. Any help? Thanks!
 A: Fix some algebraic closure $\overline{F}$ of $F$. Let $\alpha\in\overline{F}$ be a root of $T^{p^2}-a$. Then, $T^{p^2}-a$ is irreducible if and only if $[F(\alpha):F]=p^2$. But, let $\beta:=\alpha^p$. Note that since $\beta^p-a=0$, and by assumption $T^p-a$ is irreducible, we have that $[F(\beta):F]=p$. Thus, we see that $[F(\alpha):F]=p^2$ if and only if $[F(\alpha):F(\beta)]=p$. 
But, to prove this it suffices to show that $T^p-\beta$ is irreducible in $F(\beta)[x]$. But, by Capelli's theorem, it suffices to show that $T^p-\beta$ has no root in $F(\beta)$. So, suppose that $\gamma\in F(\beta)$ is such that $\gamma^p=\beta$. Note then that 
$$N_{F(\beta)/F}(\gamma)^p=N_{F(\beta)/F}(\gamma^p)=N_{F(\beta)/F}(\beta)$$
But, since the minimal polynomial of $\beta$ is $T^p-a$, we know that $N_{F(\beta)/F}$ is $(-1)^{p+1}a$. 
Thus, if $p$ is odd, then $N_{F(\beta)/F}(\beta)=a$, and thus $N_{F(\beta)/F}(\gamma)$ is a root of $T^p-a$ in $F$, which is a contradiction. 
If $p=2$, by assumption $\pm i\in F$, and so evidently $i\cdot N_{F(\beta)/F}(\gamma)$ and 
$$(i N_{F(\beta)/F}(\gamma))^2=-1 (-1)^3 a=a$$
and so $i N_{F(\beta)/F}(\gamma)$ is a root of $T^p-a$ in $F$, which is also a contradiction.
I'm not sure why we needed $F$ to contain $\zeta_p$ for $p$ odd? Can anyone see an issue with the above?
