For context: each $a_i$ is a $p$th [$p=$prime] root of some element in the field $F(a_1,..., a_{i-1})$.
The author claims that:
"if $a_i$ is a $p$th root we can assume that $F(a_1, ... , a_i)$ contains no $p$th roots of unity not in $F(a_1, ... , a_{i-1})$ unless $a_i$ itself is a $p$th root of unity. "
How could one prove this statement?
I would really appreciate any help/thoughts!
Edit: for further clarification on the meaning of $F(a_1, ..., a_i)$. The author seems to start with a generic field $F$, and adjoin a $p$th root of some element in $F$, that is, we adjoin $a_1=f_1^{1/p_1}$, $f_1∈F$ and so we get $F(a_1)$. Now we adjoin $a_2=f_2^{1/p_2}$, $f_2∈F(a_1)$ getting $F(a_1, a_2)$ and the process is repeated $i$ times, getting $F(a_1, ... , a_i)$.
Assume $p$ is prime, $K$ is an extension of your $F$ of characteristic not $p$ and $\ a \in K$ is not a root of unity and $a^{1/p} \not \in K$. Then $[K(a^{1/p}):K]=p$ (see below)
If $\zeta_n \in K(a^{1/p}),\zeta_n \not \in K$ then $K(\zeta_{n})/K$ is a non-trivial subextension, thus of degree dividing $p$, thus of degree $p$, thus $K(a^{1/p}) = K(\zeta_n)$. Take the smallest such $n$.
Find the largest $m | n$ such that $\zeta_m \in K$.
In characteristic $0$ , $\ [K(\zeta_{n}):K(\zeta_m)] = \frac{\phi(n)}{\phi(m)}$ which means $p | m $ and $n = pm$, so wlog you can set $a= \zeta_m$.
In characteristic $q$, $\ [K(\zeta_{n}):K(\zeta_m)] \ \ \ | \ \ \frac{\phi(n)}{\phi(m)}$ thus again it means $p | m $ and $n = pm$, so wlog you can set $a= \zeta_m$.
$X^p-a = \prod_{l=1}^p (X-\zeta_p^l a^{1/p})$ so that $K(\zeta_p,a^{1/p})/K(\zeta_p)$ is a Galois extension. Any element $\sigma \in Gal(K(\zeta_p,a^{1/p})/K(\zeta_p))$ is of the form $ a^{1/p} \mapsto \zeta_p^l a^{1/p}$ which has order $p$, thus $K(\zeta_p,a^{1/p})/K(\zeta_p)$ is of degree $p$ and so is $K(a^{1/p})/K$.