Why does the exponent of $\text{Gal}(F_i/F_{i-1})$ divide the exponent of $\text{Gal}(K_i/K_{i-1})$ ? We have the chain of intermediate fields $$F=K_0\leq K_1 \leq \ldots \leq K_r=K$$ 
Let $\omega$ be the $n$th primitive root of $1$. 
We consider the chain $$F\leq F(\omega)\leq F_1\leq \ldots \leq F_r=K(\omega)$$ with $F_i=K_i(\omega)$.  
We have that $F_i=F_{i-1}K_i$ and $\text{Gal}(F_i/F_{i-1})\hookrightarrow \text{Gal}(K_i/K_{i-1})$. 
Why does it hold that the exponent of $\text{Gal}(F_i/F_{i-1})$ divides the exponent of $\text{Gal}(K_i/K_{i-1})$ ? 

This has nothing to do with the specific structure of these groups - by the definition of "exponent" (for example, as the least common multiple of the orders of every element in the group), if $ H \subset G $ are finite groups, then the exponent of $ H $ always divides the exponent of $ G $...
