There is a proof of Kronecker-Weber theorem which I cannot understand (in fact I have been asking questions on several steps of the proof, and here is another one).
There is the following proposition:
Let $p\in \mathbb{N}$ be an odd prime. Then there is a unique extension $K/\mathbb{Q}$ of order $p$ such that its discriminant is a power of $p$.
The proof of the proposition is postponed and it proceeds to prove a particular case (which imply the general case by some other arguments which are not important right now):
Theorem. Let $p$ be an odd prime in $\mathbb{N}$ and let $K/\mathbb{Q}$ be a cyclic extension of degree $p^n$ whose discriminant is a power of $p$ is contained in $\mathbb{Q}\left (\zeta\right )$, where $\zeta=e^{\frac{2\pi i}{p^{n+1}}}$.
Proof. Let $L$ be the unique subfield of $\mathbb{Q}\left (\zeta\right )$ of order $p^n$. Then $KL$ is an extension whose order is a power of $p$.
$\text{Gal}\left (L/\mathbb{Q}\right )$ is cyclic, let $\tau$ be a generator and let $\sigma$ be any automorphism of $KL$ extending $\tau$. Let $F$ be the fixed field of $<\sigma >$. Then $L\cap F=\mathbb{Q}$, and there is an injection $\text{Gal}\left (LK/\mathbb{Q}\right )\to \text{Gal}\left (L/\mathbb{Q}\right )\times \text{Gal}\left (K/\mathbb{Q}\right )$ which has exponent $p^n$, so $\sigma$ has order at most $p^n$, but since it extended an element having order $p^n$, $\sigma$ has order exactly $p^n$. We will show that $[FL:F]=p^n$, hence $KL=FL$, and that $FL\subset \mathbb{Q}\left (\zeta\right )$, hence $K\subset \mathbb{Q}\left (\zeta\right )$.
If $F\neq \mathbb{Q}$ then both $L$ and $F$ contain the unique cyclic extension of degree $p$ with discriminant a power of $p$. This contradicts $L\cap F=\mathbb{Q}$. Thus $FL=L\subset \mathbb{Q}\left (\zeta\right )$ and $[FL:L]=[L:L\cap F]=[L:\mathbb{Q}]=p^n$. $\blacksquare$
I understood everything except the following: If $F\neq \mathbb{Q}$ then both $L$ and $F$ contain the unique cyclic extension of degree $p$ with discriminant a power of $p$.
Why is that true? It is true that $L$ has a subfield of order $p$ and therefore it must have discriminant a power of $p$ since it is a subextension of $\mathbb{Q}\left (\zeta\right )$, whose discriminant is a power of $p$.
If we were able to show that the same holds for $F$, then the uniqueness would give us the result, but I cannot see why it is true. I suspect that $F\subset K$ (and it trivially impies what we want) but I am not very convinced.