What does it mean to take $Q$ of some number of set of numbers?

$F(\alpha)$ denotes the minimal ring containing $F, \alpha$ and $\frac{1}{\alpha}$. $F(x_1,x_2,...,x_n)$ just means $F(x_1)(x_2)...(x_n)$.

Merge $a_i$ which are the same up to powers of $p$, and WLOG assume $a_i$ are integers. We generalize to arbitrary coefficients of the $\sqrt[p]{}$ sum. Let $t = b_1\sqrt[p]{a_1}+b_2\sqrt[p]{a_2}+\dots+b_n\sqrt[p]{a_n}$ has a minimal polynomial with degree $p^p$. Claim: No non-trivial linear relation exists between $\sqrt[p]{a_1}, \sqrt[p]{a_2}, \sqrt[p]{a_3}, \dots, \zeta$ where $\zeta$ is the $p$th primitive root of unity. Proof. Omitted. $\blacksquare$ Claim: In ${\mathbb Q}(\sqrt[p]{a_1}, \sqrt[p]{a_2}, \sqrt[p]{a_3}, \dots, \zeta)$, the map $\sigma$ that maps $\sqrt[p]{a_i}$ to $\zeta^{c_i} \cdot \sqrt[p]{a_1}$ is a well defined automorphism. Proof. Expand it all out to get the result. $\blacksquare$ Claim: $t = b_1\sqrt[p]{a_1}+b_2\sqrt[p]{a_2}+\dots+b_n\sqrt[p]{a_n}$ has a minimal polynomial with degree $p^p$. Proof. Apply the $\sigma$ automorphisms which map to giving galois conjugates of $\zeta^{c_1} b_1\sqrt[p]{a_1}+ \zeta^{c_2} b_2\sqrt[p]{a_2} + \dots + \zeta^{c_n} b_n\sqrt[p]{a_n}$. $\blacksquare$ Now, note that ${\mathbb Q}(t) \subset {\mathbb Q}(\sqrt[p]{a_1},\sqrt[p]{a_2},\dots,\sqrt[p]{a_n})$. We can compute that $[{\mathbb Q}(\sqrt[p]{a_1},\sqrt[p]{a_2},\dots,\sqrt[p]{a_n}) : {\mathbb Q}] = p^n$. However, since $t$ has a minimal polynomial of degree of $p^n$, it follows that $1, t, t^2, \dots, t^{p^n-1}$ is a basis of $[{\mathbb Q}(t) \mid {\mathbb Q}]$. It follows by tower law that \[ [{\mathbb Q}(\sqrt[p]{a_1},\sqrt[p]{a_2},\dots,\sqrt[p]{a_n}) : {\mathbb Q}(t)] = \frac{p^n}{p^n} = 1 \]which gives the result.