I don't understand why we can assume that $F$ contains any primitive m roots of unity.
Does it have anything to do with If $F \subset L$ is a radical extension, is $F(c_1,c_2,...,c_n)\subset L(c_1,c_2,...,c_n)$ radical?.
In that case are we actually trying to prove the solvability of $F(c_1,c_2,...,c_n)\subset L(c_1,c_2,...,c_n)$?
Pls see following link(I tired to prove without the assumption.): Galois group of Solvable extension is solvable. (Proof)