Galois closure of a $p$-extension is also a $p$-extension

I'm working on a problem in Dummit & 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.