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

We have $n$ distinct prime numbers $p_1,\cdots ,p_n$ and I am asked to show that $[\mathbb{Q}(\sqrt[r]{p_1},\cdots ,\sqrt[r]{p_n}):\mathbb{Q}]=r^n$ where $r\in \mathbb{N}$.

I tried to solve it by induction. The case $n=1$ is trivial. If we let

$F=\mathbb{Q}(\sqrt[r]{p_1},\cdots ,\sqrt[r]{p_n})$

$E=\mathbb{Q}(\sqrt[r]{p_1},\cdots ,\sqrt[r]{p_n},\sqrt[r]{p_{n+1}})$

$L=\mathbb{Q}(\sqrt[r]{p_2},\cdots ,\sqrt[r]{p_n})$

then by the inductive hypothesis $[F:\mathbb{Q}]=r^n$, $[L:\mathbb{Q}]=r^{n-1}$, $[F:L]=r$. We want to show that $[E:F]=r$. We know that $[E:F]\leq r$. If $[E:F]

$\prod_{i\in I\subset \left \{0,\cdots ,r-1\right \}}X-\sqrt[r]{p_{n+1}}\zeta_r ^i$ where $\left |I\right |=m

Therefore $\sqrt[r]{p_{n+1}^m}=\sum_{i=0}^{r-1}a_i\left (\sqrt[r]{p_1}\right )^i$ where $a_i\in L$.

I don't know how to continue, I don't even know if I am doing the right thing.

Is it true that the trace over $F/L$ of $\sqrt[r]{p_{1}^i}$ equals $0$ for every $i\neq 0$? Because in that case taking trace over $F/L$ we would obtain $0=ra_0$, thereby $a_0=0$ and I think that could be helpful.