(All fields below are subfields of the field $\mathbb{C}$).
I am looking for a proof (or disproof) of the following statement:
Let $L:K$ be a normal extension and $K = N_{0} \subset N_{1} \subset N_{2} \ldots \subset N_{n} = L$, ($n \geq 1$)
be a normal soluble tower, that is $N_{i} : N_{i-1}$ is normal and $Gal(N_{i} : N_{i-1})$ is soluble, $i=1,2, \ldots, n$.
Then $Gal(L:K)$ is soluble.
A proof sketch. By induction on $n$.
If $n = 1$ then $K = N_{0}$ and $L=N_{1}$. But it is given that $N_{1} : N_{0}$ is soluble.
If $n > 1$ then consider the last 3 fields of the tower. That is,
$N_{n-2} \subset N_{n-1} \subset N_{n} = L$.
The extension $L:N_{n-2}$ is normal (because $L:K$ is normal and $N_{n-2}$ is an intermediate field).
By Galois correspondence, $Gal(N_{n-1}:N_{n-2}) \cong Gal(L:N_{n-2})/Gal(L:N_{n-1})$.
Therefore, $Gal(L:N_{n-2})$ is soluble. (The theorem "If H and G/H are soluble, then G is soluble" was used).
Then $K = N_{0} \subset N_{1} \subset N_{2} \ldots N_{n-2} \subset N_{n} = L$ is a normal soluble tower of a lesser length.
By induction, $Gal(L:K)$ is soluble.
Any references to the literature are welcome.