Prove that $[\mathbb{Q}(\sqrt[r]{p_1},\cdots ,\sqrt[r]{p_n}):\mathbb{Q}]=r^n$

This question has been asked many times for the particular case $r=2$. The central technical difficulty consists in translating a condition of additive linear independence into a multiplicative one. The most convenient way is to use Kummer theory under certain hypotheses which render the problem easier than in the general references given by @Bill Dubuque. For example:

Fix $r\ge2$. Let $K$ be the cyclotomic field $\mathbf Q(\zeta_r)$ and let us prove by induction the property $(P_m)$ that $[K(\sqrt [r] {p_1} , ... , \sqrt [r]{p_m}): K]=r^m$ for all $m$'s and all primes $p_j$'s not dividing $r$, i.e. unramified in $K$. For $m=1$, the Eisenstein criterion shows that $[\mathbf Q(\sqrt[r] {p_1}):\mathbf Q]= r$, hence, by linear disjointness (which follows from the non ramification condition), $(P_1)$ holds. Suppose that $(P_{n})$ holds. Then, by Kummer theory, the subgroup $B_n$ of $K^{*}/K^{*r}$ generated by the classes of $p_1,...,p_n$ mod $K^{*r}$ is isomorphic to $(\mathbf Z/r)^{n}$ in additive notation. Since $(P_{n+1})$ is equivalent to the linear disjointness of $K(\sqrt [r] {p_1} , ... , \sqrt[r]{p_n})$ and $ K(\sqrt[r] p_{n+1})$, it amounts to showing that for any $a_{n+1} < r$, the class of $p_{n+1}^{a_{n+1}}$ cannot belong to $B_n$. Suppose the contrary, i.e. that that we have an equation $p_{n+1}^{a_{n+1}} = p_{1}^{a_1}...p_{n}^{a_n}x^r$, where $a_j < r$ and $x\in K^{*}$, or, after clearing denominators, an equality of ideals $(p_{n+1}^{a_{n+1}}).I^r = (p_{1}^{a_1}...p_{n}^{a_n}).J^r$ (¤), where $I$ and $J$ are two coprime integral ideals. To simplify notations, write $p$ for any of the $p_j$'s. Since $p$ is unramified, the principal ideal $(p)$ is prime or splits as a product $P_1...P_s$ of $s$ distinct prime ideals. The participation of $(p)$ or of any such $P_i$ to the equation (¤) is of the form $(p)^a$ or $P_i^{a}$, with $a<r$. Impossible,by the uniqueness of the prime decomposition of ideals in a Dedekind ring. Thus $(P_m)$ holds for $K$, hence, by linear disjointness, the analogous property holds for $\mathbf Q$.

Question: The non ramification hypothesis is not needed when $r=2$. When $r>2$ what happens for the divisors of $r$ ?