I am trying to prove the following quoted statement.
Theorem 1. Let $r_1,...,r_n$ be real algebraic integers over $\Bbb Q$ with minimal polynomials $x^{d_1}-a_1,...,x^{d_n}-a_n$, respectively. Then the equality $$\left[\Bbb Q[r_1,...,r_n]:\Bbb Q\right]=\prod_{i=1}^nd_i$$holds if and only if for all $s_i\in\Bbb Z$, $$\prod_{i=1}^nr_i^{s_i}\in \Bbb Q$$implies $s_i\equiv 0$ modulo $d_i$ for all $1\leq i\leq n$.
I will sketch a proof that is "almost done", requiring some technical modifications.
Let $m=\operatorname{lcm}(d_1,...,d_n)$, let $\omega$ be a primitive $m^{th}$ root of unity and let $\mu_m(\Bbb Q)$ to be the group of all $m^{th}$ root of unity over $\Bbb Q$. For clarity set $\Bbb Q_{\omega}:=\Bbb Q[\omega]$ and $E=\Bbb Q_{\omega}[r_1,...,r_n]$. Define the large Kummer group $$\operatorname{KUM}\left(E/\Bbb Q_{\omega}\right)=\{r\in E:r^m\in\Bbb Q_{\omega}^{\times}\}$$and define the small Kummer group to be the quotient $$\operatorname{kum}\left(E/\Bbb Q_{\omega}\right)=\operatorname{KUM}\left(E/\Bbb Q_{\omega}\right)/\Bbb Q_{\omega}^\times.$$Since $\Bbb Q_{\omega}^{\times}$ contains all $m^{th}$ roots of unity, the group $\operatorname{Gal}\left(E/\Bbb Q_{\omega}\right)$ is abelian and the map $$\operatorname{Gal}\left(E/\Bbb Q_{\omega}\right)\times \operatorname{kum}(E/\Bbb Q_{\omega})\to\mu_m(\Bbb Q)$$defined by $$(\sigma,r)\mapsto\frac{\sigma(r)}{r}$$is bilinear as $\Bbb Z$-modules. Moreover it is non-degenerate by the fundamental theorem of Galois theory and the fact that a Galois automorphism is determined by the values of the generators. Since both $\operatorname{Gal}\left(E/\Bbb Q_{\omega}\right)$ and $\operatorname{kum}(E/\Bbb Q_{\omega})$ are finite abelian and $\mu_m(\Bbb Q)$ is finite cyclic, It is a group theory exercise that then $$\operatorname{Gal}\left(E/\Bbb Q_{\omega}\right)\simeq \operatorname{kum}(E/\Bbb Q_{\omega}).$$
The case where $m=d_1=...=d_n=2$in Theorem 1 is proved, since $\Bbb Q_{\omega}=\Bbb Q$ and the list of elements of them form $\prod_{i=1}^nr_i^{s_i}$ for $0\leq s_i\leq d_i-1$ gives $\prod_{i=1}^nd_i$ distinct cosets in $\operatorname{kum}(E/\Bbb Q)$ while spanning the extension $E/\Bbb Q$. So one must have $$\prod_{i=1}^nd_i\leq \operatorname{kum}(E/\Bbb Q)=\operatorname{Gal}(E/\Bbb Q)\leq \prod_{i=1}^nd_i$$as needed.
However, the general case of Theorem 1 does not follow directly from the isomorphism between the small kummer group and the Galois group. Consider the case where $r_1=\sqrt{3},r_2=\sqrt[4]{5},r_3=\sqrt[3]{7}$. Then the extension $$\Bbb Q_{\omega}\left[\sqrt{3},\sqrt[4]{5},\sqrt[3]{7}\right]=\Bbb Q\left[i,\sqrt 3\right]\left[\sqrt{3},\sqrt[4]{5},\sqrt[3]{7}\right]=\Bbb Q\left[i,\sqrt 3,\sqrt[4]{5},\sqrt[3]{7}\right].$$The products of the $r_i$'s in the small kummer group are NOT distinct as $\sqrt 3\in\Bbb Q_{\omega}$, so gives the same cosets if multiplied by another $r_i$. But I believe Theorem 1. in this case is indeed correct, while one needs some technical modifications.
Is there an elegant way to resolve this?
This answer is to prove the following theorem, which implies the theorem in the question by looking at linear independence.
Let $E_n=\Bbb Q\left[\sqrt[n]{p_1},...,\sqrt[n]{p_m}\right]$. As mentioned in the question the equality $$\left[E_2:\Bbb Q\right]=\left[\Bbb Q\left[\sqrt{p_1},...,\sqrt{p_m}\right]:\Bbb Q\right]=2^m$$has been established. Although the cyclotomic extension of $\Bbb Q$ might contain square roots (in fact every square root of a positive integer is contained in some cyclotomic extension of $\Bbb Q$,) we could build upon the obtained equality $[E_2:\Bbb Q]=2^m$ and show that higher roots does not appear to be contained in a cyclotomic extension of $\Bbb Q$.
The first lemma is some well-known facts that we are going to use.
In the sense of Kronecker–Weber theorem, the following lemma says that higher roots of positive integers are not contained in a cyclotomic field.
Proof. The case of either $n$ or $k$ being $1$ is obvious. Assume now $n,k>1$.
If $\left(\sqrt[n]{k}\right)^2\in\Bbb Q$, write $\sqrt[n]{k}=\sqrt{a}$ for some $a\in\Bbb Q_{>0}$. Then $\Bbb Q\subset \Bbb Q\left[\omega_n,\sqrt{a}\right]$ must be abelian as the composition of abelian Galois extensions $\Bbb Q\subset \Bbb Q[\omega_n]$ and $\Bbb Q\subset\Bbb Q\left[\sqrt{a}\right]$.
Conversely assume $\left(\sqrt[n]{k}\right)^2\notin\Bbb Q$, and write $\sqrt[n]{k}=\sqrt[r]{t}$ with the smallest positive integer $r$ possible. Clearly then $r|n$ and $r>2$. Note that $\Bbb Q\subset\Bbb Q\left[\omega_n,\sqrt[r]{t}\right]=\Bbb Q\left[\omega_n,\sqrt[n]{k}\right]$ is Galois as $\Bbb Q\left[\omega_n,\sqrt[r]{t}\right]$ is the splitting field of $x^n-k$ over $\Bbb Q$.
Case 1. If $r$ contains an odd prime factor $p$, then $\sqrt[p]{t}\notin\Bbb Q$ as well and$$\Bbb Q\subset \Bbb Q\left[\omega_p,\sqrt[p]{t}\right]\subset\Bbb Q\left[\omega_n,\sqrt[r]{t}\right],$$where all extensions above are Galois. To prove $\Bbb Q\subset\Bbb Q\left[\omega_n,\sqrt[r]{t}\right]$ is non-abelian it suffices to prove $\Bbb Q\subset \Bbb Q\left[\omega_p,\sqrt[p]{t}\right]$ is non-abelian. By the irreducibility of cyclotomic polynomials and Lemma 1.$(i)$, there are $\sigma,\tau\in\operatorname{Gal}\left(\Bbb Q\left[\omega_p,\sqrt[p]{t}\right]/\Bbb Q\right)$ such that $$\sigma(\omega_p)=\omega_p,\tau(\omega_p)=\omega_p^2~\text{ and }~\sigma\left(\sqrt[p]{t}\right)=\omega_p\sqrt[p]{t},\tau\left(\sqrt[p]{t}\right)=\sqrt[p]{t}.$$But $\sigma$ and $\tau$ does not commute on $\sqrt[p]{t}$ as $$\sigma\tau\left(\sqrt[p]{t}\right)=\omega_p\sqrt[p]{t}~\text{ and }~\tau\sigma\left(\sqrt[p]{t}\right)=\omega_p^2\sqrt[p]{t},$$showing that $\Bbb Q\subset \Bbb Q\left[\omega_p,\sqrt[p]{t}\right]$ is non-abelian.
Case 2. If $r$ is a power of $2$, then $4|r$, $\sqrt[4]{t}\notin\Bbb Q$ and $$\Bbb Q\subset \Bbb Q\left[\omega_4,\sqrt[4]{t}\right]\subset\Bbb Q\left[\omega_n,\sqrt[r]{t}\right]$$where all extensions above are Galois. To prove $\Bbb Q\subset\Bbb Q\left[\omega_n,\sqrt[r]{t}\right]$ is non-abelian it suffices to prove $\Bbb Q\subset \Bbb Q\left[\omega_4,\sqrt[4]{t}\right]$ is non-abelian. By the irreducibility of cyclotomic polynomials and Lemma 1.$(ii)$, there are $\sigma,\tau\in\operatorname{Gal}\left(\Bbb Q\left[\omega_4,\sqrt[4]{t}\right]/\Bbb Q\right)$ such that $$\sigma(\omega_4)=\omega_4,\tau(\omega_4)=\omega_p^3~\text{ and }~\sigma\left(\sqrt[4]{t}\right)=\omega_4\sqrt[4]{t},\tau\left(\sqrt[4]{t}\right)=\sqrt[4]{t}.$$But $\sigma$ and $\tau$ does not commute on $\sqrt[4]{t}$ as $$\sigma\tau\left(\sqrt[4]{t}\right)=\omega_4\sqrt[4]{t}~\text{ and }~\tau\sigma\left(\sqrt[4]{t}\right)=\omega_4^3\sqrt[4]{t},$$showing that $\Bbb Q\subset \Bbb Q\left[\omega_4,\sqrt[4]{t}\right]$ is non-abelian.$\square$
Now clearly $$n^m\geq [E_n:\Bbb Q]\geq[E_{2n}:E_2]\geq [E_{2n}[\omega_{2n}]:E_{2}[\omega_{2n}]]$$by the diagram below.
By Kummer Theory it suffices to show the estimate $$\#\operatorname{kum}\left(E_{2n}[\omega_{2n}]/E_2[\omega_{2n}]\right)\geq n^m.$$
Proof. Define the obvious generating set $$B_{2n}^2=\Bigl\{\sqrt[2n]{p_1^{s_1}...p_m^{s_m}}:0\leq s_i\leq n-1\Bigr\}$$for $E_{2n}[\omega_{2n}]$ over $E_2[\omega_{2n}]$. I claim $B_{2n}^2$ gives distinct cosets in $\operatorname{kum}\left(E_{2n}[\omega_{2n}]/E_2[\omega_{2n}]\right)\geq n^m$.
Let us fix $r\in B_{2n}^2$. It remains show $r\in \left(E_2[\omega_{2n}]\right)^\times$ implies $r=1$.
The extensions $\Bbb Q\subset E_2$ and $\Bbb Q\subset \Bbb Q[\omega_{2n}]$ are Galois and abelian, so should their composition $\Bbb Q\subset E_2\Bbb Q[\omega_{2n}]$. If $r\in \left(E_2[\omega_{2n}]\right)^\times$ then $\Bbb Q\subset\Bbb Q\left[r,\omega_{2n}\right]\subset E_2\Bbb Q[\omega_{2n}]$ so that the Galois group of $\Bbb Q\subset\Bbb Q\left[r,\omega_{2n}\right]$ is a quotient of that of $\Bbb Q\subset E_2\Bbb Q[\omega_{2n}]$, which must be abelian.
Assume $r\neq 1$. Clearly $r^2\notin\Bbb Q$. Lemma 1 shows $\Bbb Q\subset\Bbb Q\left[r,\omega_{2n}\right]$ is nonabelian, contradiction.$\square$
We have proved a stronger result that $$[E_n:\Bbb Q]=[E_{2n}:E_2]= [E_{2n}[\omega_{2n}]:E_{2}[\omega_{2n}]]=n^m.$$