I am trying to prove the following:
Let $K=\mathbb{Q}(\sqrt{p_1},\ldots,\sqrt{p_n})$. Show that $K/\mathbb{Q}$ is Galois with Galois group $(\mathbb{Z}/2\mathbb{Z})^n$.
I have attached my proof below not as a means of verifying it, only as a means of making my question as clear as possible.
My question is:
Is it correct to say that that $\tau_i$'s generate $\operatorname{Gal}(K/\mathbb{Q})$, and therefore that we can decompose elements of $\operatorname{Gal}(K/\mathbb{Q})$ into compositions of these functions.
My proof is as follows:
To show $K/\mathbb{Q}$ is Galois we need to show it is the splitting field of a separable polynomial. The field $\mathbb{Q}(\sqrt{p_1},\ldots,\sqrt{p_n})$ is obtained by an $n$ step process where step $i$ consists of adjoining $\sqrt{p_i}$ to the field $Q_{i-1}=\mathbb{Q}(\sqrt{p_1},\ldots,\sqrt{p_{i-1}})$ and $Q_0$ is defined to be $\mathbb{Q}$. This gives the following tower of extensions
$$\mathbb{Q}\subset \mathbb{Q}(\sqrt{p_1})\subset \mathbb{Q}(\sqrt{p_1},\sqrt{p_2})\subset \cdots\subset \mathbb{Q}(\sqrt{p_1},\ldots,\sqrt{p_n}).$$
Note that $Q_i\neq Q_{i+1}$, because $\sqrt{p_{i+1}} \not \in Q_{i}$ for all $i$; the fact the $\sqrt{p_1},\ldots,\sqrt{p_n}$ are distinct primes insures this. Since none of the $p_j$ are squares (because they are prime) we know $\sqrt{p_j} \not \in \mathbb{Q}$. Furthermore, the distinctness of the $p_j$ insure that $\sqrt{p_i} \neq \sqrt{p_j}$ for all $i \neq j$.
The minimal polynomial of the extension $Q_{i+1}/Q_i$ is $f_{i+1}(X)=X^2-p_{i+1}$, which is separable with roots $\pm\sqrt{p_{i+1}}$. From this we see that the polynomial we are concerned with is
$$f(X)=f_1(X)\cdots f_n(X)=(X^2-p_1)\cdots(X^2-p_n),$$
also a separable polynomial. $K$ is a splitting field for this separable polynomial over $\mathbb{Q}$, and thus $K/\mathbb{Q}$ is Galois. From the tower of fields above we see they $[K:\mathbb{Q}]=2^n$, since the degree of each $Q_i/Q_{i-1}$ is 2. From this we can deduce that $\operatorname{Gal}(K/\mathbb{Q})=2^n$. As with before, $\operatorname{Gal}(K/\mathbb{Q})$ is determined by its action on the roots of $f(X)$, which are
$$\{\pm \sqrt{p_1},\ldots,\pm \sqrt{p_n}\},$$
or, more correctly, on the roots of the minimal polynomial of each of these roots. Consider the map
\begin{equation*}
\tau_{(a_1,\ldots,a_n)}= \begin{cases}
\sqrt{p_1}& \mapsto \pm \sqrt{p_1},\\
\sqrt{p_2}& \mapsto \pm \sqrt{p_2},\\
& \vdots\\
\sqrt{p_n}& \mapsto \pm \sqrt{p_n},
\end{cases}
\end{equation*}
where $a_i\in \{0,1\}$ and $a_i=0$ means $\sqrt{p_i} \mapsto \sqrt{p_i}$ and $a_i=1$ means $\sqrt{p_i} \mapsto \sqrt{p_i}$.
This gives $2n$ possible maps which take roots of the $f_i(X)$ to roots of $f_i(X)$ (note that these are all such maps as well). Since $|\operatorname{Gal}(K/\mathbb{Q})|=2^n$ all of these maps must be automorphisms of $K$.
What remains to be seen is that $\operatorname{Gal}(K/\mathbb{Q}) \cong (\mathbb{Z}/2\mathbb{Z})^n$. We can decompose $\tau_{(a_1,\ldots,a_n)}$ into a composition of the maps $\tau_i:K\rightarrow K$ defined by
\begin{equation*}
\tau_i = \begin{cases}
\tau_i\left(\sqrt{p_j}\right) = -\sqrt{p_i} & \text{for } i=j,\\
\tau_i\left(\sqrt{p_j}\right) = \sqrt{p_j} & \text{for } i\neq j,
\end{cases}
\end{equation*}
where $\tau_{(a_1,\ldots,a_n)}=\tau_1^{a_1}\circ \cdots \circ \tau_n^{a_n}$. We see that every element of $\operatorname{Gal}(K/\mathbb{Q})$ can be obtained through the composition of $\tau_i$'s and therefore $\{\tau_1,\ldots,\tau_n\}$ generates $\operatorname{Gal}(K/\mathbb{Q})$.
Let $\tau \in \operatorname{Gal}(K/\mathbb{Q})$ such that $\tau_1^{a_1}\circ \cdots \circ \tau_n^{a_n}$. Define the map $\chi: \operatorname{Gal}(K/\mathbb{Q}) \rightarrow \{\pm 1\}^n$ by first defining the action of $\chi$ on each $\tau_i$ by
$$\chi(\tau_i) \mapsto \tau_i(\sqrt{p_i})/\sqrt{p_i},$$
and then defining
\begin{align*}
\chi(\tau)&=\chi(\tau_1^{a_1}\circ \cdots \circ \tau_n^{a_n})\\
&=(\chi(\tau_1^{a_1}),\ldots,\chi(\tau_n^{a_n})).
\end{align*}
Once we show this map is injective we are done, as any injective map between finite sets of the same cardinality is surjective. Assume $(a_1,\ldots,a_n)=(b_1,\ldots,b_n)$. Then clearly $\tau_1^{a_1}\circ \cdots\circ \tau_n^{a_n}=\tau_1^{b_1}\circ \cdots \circ \tau_n^{b_n}$, since $a_i=b_i$ for all $i$.
Now, $\{\pm 1\}$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}$, as the map $\varphi: \{\pm 1\} \rightarrow \mathbb{Z}/2\mathbb{Z}$ by $\varphi(-1)=1$ and $\varphi(1)=0$ is an injective homormorphism. Therefore
$$\operatorname{Gal}(K/\mathbb{Q})\cong (\mathbb{Z}/2\mathbb{Z})^n.$$
Attempted proof of Martin Brandenburg's Claim:
Claim: For all $i<n$ and $i<k_1<\ldots<k_s\leq n$, $\sqrt{p_{k_1}\ldots p_{k_s}} \not \in Q_i$.
Proceed by induction on $n$. If $i=0$, then we must show that $\sqrt{p_{k_1}\ldots p_{k_s}} \not \in \mathbb{Q}$, where $0<k_1<\ldots<k_s\leq n$. This is equivalent to showing that $p_{k_1}\ldots p_{k_s}$ is not a square, which is clear since $p_{k_1}, \ldots ,p_{k_s}$ are prime numbers. Assume the result is true for all $i<n-1$ and consider $i=n-1$. We need to show $\sqrt{p_n}\not \in Q_{n-1}$. Assume the contrary, that $\sqrt{p_n} \in Q_{n-1}$. Then
$$\sqrt{p_n}=\sum_{(\alpha_1,…,\alpha_{n-1})} a_{(\alpha_1,…,\alpha_{n-1})} \sqrt{{p_1}^{\alpha_1} \ldots p_{n-1}^{\alpha_{n-1}}}, \quad \dagger$$
where $\alpha_i \in \{0,1\}$ and $a_{(\alpha_1,…,\alpha_{n-1})}\in \mathbb{Q}$. Multiplying both sides of $(\dagger)$ by $\sqrt{p_n}$ gives
$$p_n=\sum_{(\alpha_1,…,\alpha_{n-1})} a_{(\alpha_1,…,\alpha_{n-1})} \sqrt{{p_1}^{\alpha_1} \ldots p_{n-1}^{\alpha_{n-1}}p_n},$$
contradicting the fact $\sqrt{{p_1}^{\alpha_1} \ldots p_{n-1}^{\alpha_{n-1}}p_n} \not \in \mathbb{Q}$. Therefore $\sqrt{p_n} \not \in Q_{n-1}$.
Best Answer
The crucial point is: It is not trivial at all to see that $\mathbb{Q}(\sqrt{p_1},\dotsc,\sqrt{p_n})$ has degree exactly $2^n$ (only $\leq 2^n$ is clear) over $\mathbb{Q}$, or equivalently, that $p_i \notin \mathbb{Q}(\sqrt{p_1},\dotsc,\sqrt{p_{i-1}})$ for all $i \leq n$.
Here is what you can do: Let $m_1,\dotsc,m_r$ be coprime and square free numbers $>1$ (for example distinct primes). Then I claim that $K_r := \mathbb{Q}(\sqrt{m_1},\dotsc,\sqrt{m_r})$ has degree $2^r$ over $\mathbb{Q}$.
It is enough to prove $\sqrt{m_{i+1}} \notin K_i$ for all $i < r$. But it turns out one should show something stronger: For $i<r$ and $i < j_1 < \dotsc < j_s \leq r$ we have $\sqrt{m_{j_1} \cdot \dotsc \cdot m_{j_s}} \notin K_i$. Now this can be shown by induction on $i$; I leave it for you as an exercise.
Now the rest is easy: $K_r$ is the splitting field of $(x^2-m_1) \cdot \cdots \cdot (x^2-m_r)$, hence normal, and everything in characteristic $0$ is separable. Thus $K_r$ is Galois over $\mathbb{Q}$; let $G$ be the Galois group. Since the conjugates of $\sqrt{m_i}$ are $\pm \sqrt{m_i}$, there is a homomorphism $\alpha : G \to \{\pm 1\}^r$ given by $\alpha(g) \sqrt{m_i} = g(\sqrt{m_i})$. Clearly it is injective. Since $G$ has order $2^r$, it is also surjective. Thus we have an isomorphism $G \cong \{\pm 1\}^r$. Explicitly, the Galois automorphisms are exactly as you've described them, they are given by $\sqrt{m_i} \mapsto \varepsilon_i \sqrt{m_i}$, where $\varepsilon_i = \pm 1$.
Remark: There are shorter proofs, using Kummer Theory.