For $n\ge 0$ let $E_n=\Bbb Q(\sqrt {p_1},\ldots, \sqrt{p_n})$ be the smallest field extension of $\Bbb Q$ containing $\sqrt{p_k}$ for $1\le k\le n$.
Claim. For every $(\epsilon_1,\ldots, \epsilon_n)\in\{-1,1\}^n$, there is an automorphism $\phi$ of $E_n$ with $\phi(\sqrt{p_i})=\epsilon_i\sqrt{p_i}$, $1\le i\le n$.
Proof. [By induction]. The claim is vacuously true for $n=0$.
Let $n\ge 0$ and assume $\sqrt{p_{n+1}}\in E_{n}$. Clearly, $E_{n}$ is spanned as a $\Bbb Q$-vector space by all products of some of the $\sqrt{p_i}$ (including the empty product, $1$). Thus we can write
$$\tag1\sqrt{p_{n+1}}=\sum_{S\subseteq\{1,\ldots,n\}}q_S\prod_{i\in S}\sqrt{p_i}$$
with $q_S\in \Bbb Q$. Among all such representations, pick one with the minimal number of non-zero coefficients. Assume there are at least two non-zero coefficients $q_A, q_B$.
Pick $k\in A\mathop{\Delta}B$. By induction hypothesis, there is an automorphism $\phi$ of $E_n$ that maps $\sqrt{p_k}\mapsto-\sqrt{p_k}$ and for $i\ne k$ maps $\sqrt{p_i}\mapsto\sqrt{p_i}$. Also, $\phi(\sqrt{p_{n+1}})=\pm\sqrt{p_{n+1}}$ so that one of $\frac{\sqrt{p_{n+1}}\pm \phi(\sqrt{p_{n+1}})}2$ equals $\sqrt{p_{n+1}}$. This way, we obtain another representation of $\sqrt{p_{n+1}}$ of the form $(1)$, but with less non-zero coefficients because at least either $q_A$ or $q_B$ (depending on the "$\pm$")
is replaced with a $0$. From this contradiction, we conclude that in a minimal representation, at most one $q_S$ in $(1)$ is non-zero. Thus $\sqrt{p_{n+1}}=q_S\prod_{i\in S}\sqrt{p_i}$, contradicting the irrationality of $\prod_{i\in S}\sqrt{p_i}\sqrt{p_{n+1}}$.
We conclude that $\sqrt{p_{n+1}}\notin E_n$, hence $E_{n+1}$ is a quadratic extension of $E_n$. Also, $\sqrt{p_{n+1}}\mapsto -\sqrt{p_{n+1}}$ is an automorphism of $E_{n+1}$ over $E_n$ and the claim follows for $n+1$. $\square$
