Suppose $K/F$ is a Galois extension of degree $p^m$. Then there is a chain of extensions $F \subseteq F_1 \subseteq \cdots F_m = K$ each of degree $p$

Finite $p$-groups are super solvable: in fact, you can find a chain of subgroups $G_0=\{1_G\}\subset \cdots G_i\subset G_{i+1}\subset\cdots \subset G_m=G$ such that each subgroup $G_i$ has order $p^i$ and is normal in $G$ (not only in $G_{i+1}$ !)

The proof is by induction on $m$. The case $m=0$ is clear. Now if $G$ has order $p^{m+1}$, it has non trivial center, so by Cauchy theorem, we may pick $z\in Z(G)$ of order $p$. Remember that a (normal) subgroup of the quotient may be written as $H/\langle z\rangle$, where $H$ is a (normal) subgroup of $G$ containing $\langle z\rangle $, and that $H$ is unique.

Now apply induction on $G/\langle z\rangle$ to find an appriopriate chain of normal subgroups, and lift it via $G\to G/\langle z\rangle$ to get a chain of subgroups ending on the left by $\langle z\rangle$. To conclude, add the trivial subroup on the left and you are done.