Galois closure of a $p$-extension is also a $p$-extension I'm working on a problem in Dummit &amp; Foote and I'm quite stumped. The problem reads:


  Let $p$ be a prime and let $F$ be a field. Let $K$ be a Galois extension of $F$ whose
  Galois group is a $p$-group (i.e., the degree $[K:F]$ is a power of $p$). Such an 
  extension is called a $p$-extension (note that $p$-extensions are Galois by 
  definition).  
  
  Let $L$ be a $p$-extension of $K$. Prove that the Galois closure of $L$ over $F$ is a $p$-extension of $F$.


This is what I've done so far:


  Using the tower law we can readily show that $L$ is a $p$ extension of $F$ so we have $[L:F]=p^{\ell}$ for some integer $\ell$.  Then if $M$ is the Galois closure of $L$ over $F$ then $$[M:F]=[M:L][L:F]$$ and therefore $[M:F]=p^{\ell}n$ for some integer $n$ that is not divisible by $p$.  So $[M:L]=n$.


From here it seems like I want to show that either $n=1$ or that $n$ is in fact a power of $p$. I just don't see how to proceed. I've considered using the Sylow Theorems, but I'm not sure how that would really work.  I also realize that this statement depends on $K$ being Galois over $F$ but I can't figure out how to take advantage of that.
First, $L/F$ is a finite and separable extension, thus by the primitive element theorem $L = F(\alpha)$ for some $\alpha \in L$. Let $m_\alpha \in F[X]$ be the minimal polynomial of $\alpha$.  Since $K/F$ is Galois, over $K$, $m_\alpha$ splits into irreducible factors of the same degree, $d$ say. In Dummit-Foote this last statement is Exercise 14.4.4, the exercise right before the one in the question.  Since $d$ divides $p^\ell = [L:F]$, it is a power of $p$ itself.  This proves 

Claim 1.  For any root $\beta \in M$ of $m_\alpha$, $K(\beta)/K$ has degree a power of $p$.  

Now the Galois closure $M$ is the composite field of the $K(\beta_i)$ where $\beta_i$ runs through the roots of $m_\alpha$.  Reason: The composite field is the splitting field of $m_\alpha$, so it is Galois and contains $M$.  On the other hand $M$ contains the splitting field of $m_\alpha$ because it contains the root $\alpha$.

Claim 2. $K(\beta)/K$ is Galois for every root $\beta$ of $m_\alpha$.

Proof. $L/K = K(\alpha)/K$ is Galois.  Let $\phi : K(\alpha) \to K(\beta)$ be the isomorphism induced by $\alpha \mapsto \beta$ and the identity on $K$. Then $\text{Aut}(K(\alpha)/K) \cong \text{Aut}(K(\beta)/K)$ via $\sigma \mapsto \phi\sigma\phi^{-1}$.

Finally, the Galois group of $M$ (the composite of the $K(\beta)$) over $K$, is a subgroup of the direct product of the Galois groups of $K(\beta)/K$, which are all $p$-groups by Claim 1.  Therefore it is a p-group by Dummit-Foote Proposition 14.12.  Therefore $[M:K]$ is a power of $p$ and since $[K:F]$ is a power of $p$ and the same follows for $[M:F]$ by the tower formula.
We denote by $G(K/F)$ the Galois group of a Galois extension $K/F$.
We need the following lemma.

Lemma
Let $F$ be a field, $\Omega$ its algebraic closure.
Let $K_1$ and $K_2$ be subfields of $\Omega$ each of which is a Galois extension of $F$.
Then $K_1K_2$ is a Galois extension of $F$ and $G(K_1K_2/F)$ is isomorphic to a subgroup of 
$G(K_1/F)\times G(K_2/F)$.

Proof.
Let $\sigma$ be an automorphism of $\Omega/F$.
Then $\sigma(K_1K_2) = \sigma(K_1)\sigma(K_2) = K_1K_2$.
Hence $K_1K_2/F$ is a normal extension.
Clearly it is a separable extension.
Hence $K_1K_2/F$ is a Galois extension.
We define a homomorphism $\psi\colon G(K_1K_2/F) \rightarrow G(K_1/F)\times G(K_2/F)$
by $\psi(\sigma) = (\sigma\mid K_1, \sigma\mid K_2)$.
Since it is clearly injective, we are done.

Now let $F$ be a field, $\Omega$ its algebraic closure.
Let $F \subset K \subset L \subset \Omega$ be a tower of fields such that $K/F$ and $L/K$ are $p$-extensions. Let $M/F$ be the Galois closure of $L/F$ in $\Omega$.
Then $M = \sigma_1(L)\cdots \sigma_n(L)$, where $\sigma_i(L), i = 1,\cdots, n$ are all the distinct conjugates of $L$ over $F$ in $\Omega$.
Since each $\sigma_i(L)$ is a Galois extension over $K$, $G(M/K)$ is isomorphic to a subgroup of
$G(\sigma_1(L)/K)\times \cdots \times G(\sigma_n(L)/K)$ by the lemma. Since each $G(\sigma_i(L)/K)$ is a $p$-group, $G(M/K)$ is a $p$-group. Since $K/F$ is a $p$-extension, $(M : F)$ is a power of $p$.
Hence $M/F$ is a $p$-extension and we are done.