Proving that $\left(\mathbb Q[\sqrt p_1,\dots,\sqrt p_n]:\mathbb Q\right)=2^n$ for distinct primes $p_i$.
HINT $\ $ An inductive proof follows easily from this
LEMMA $\rm\ \ [K(\sqrt{a},\sqrt{b}) : K] = 4\ $ if $\rm\ \sqrt{a},\ \sqrt{b},\ \sqrt{a\:b}\ $ all are not in $\rm\:K\:$ and $\rm\: 2 \ne 0\:$ in $\rm\:K\:.$
Proof $\ \ $ Let $\rm\ L = K(\sqrt{b})\:.\:$ Then $\rm\: [L:K] = 2\:$ via $\rm\:\sqrt{b} \not\in K\:,\:$ so it is sufficient to prove $\rm\: [L(\sqrt{a}):L] = 2\:.\:$ It fails only if $\rm\:\sqrt{a} \in L = K(\sqrt{b})\ $ and then $\rm\ \sqrt{a}\ =\ r + s\ \sqrt{b}\ $ for $\rm\ r,s\in K\:.\:$ But that is impossible since squaring yields $\rm(1):\ \ a\ =\ r^2 + b\ s^2 + 2\:r\:s\ \sqrt{b}\:,\: $ which contradicts hypotheses as follows:
$\rm\qquad\qquad rs \ne 0\ \ \Rightarrow\ \ \sqrt{b}\ \in\ K\ \ $ by solving $(1)$ for $\rm\sqrt{b}\:,\:$ using $\rm\:2 \ne 0$
$\rm\qquad\qquad\ s = 0\ \ \Rightarrow\ \ \ \sqrt{a}\ \in\ K\ \ $ via $\rm\ \sqrt{a}\ =\ r \in K$
$\rm\qquad\qquad\ r = 0\ \ \Rightarrow\ \ \sqrt{a\:b}\in K\ \ $ via $\rm\ \sqrt{a}\ =\ s\ \sqrt{b}\:,\: \ $times $\rm\:\sqrt{b}\quad\quad$ QED
Using the above as the inductive step one easily proves the following result of Besicovic.
THEOREM $\ $ Let $\rm\:Q\:$ be a field with $2 \ne 0\:,\:$ and $\rm\ L = Q(S)\ $ be an extension of $\rm\:Q\:$ generated by $\rm\: n\:$ square roots $\rm\ S = \{ \sqrt{a}, \sqrt{b},\ldots \}$ of elts $\rm\ a,\:b,\:\ldots \in Q\:.\:$ If every nonempty subset of $\rm\:S\:$ has product not in $\rm\:Q\:$ then each successive adjunction $\rm\ Q(\sqrt{a}),\ Q(\sqrt{a},\:\sqrt{b}),\:\ldots$ doubles the degree over $\rm\:Q\:,\:$ so, in total, $\rm\: [L:Q] \ =\ 2^n.\:$ Hence the $\rm2^n$ subproducts of the product of $\rm\:S\:$ comprise a basis of $\rm L$ over $\rm\:Q\:.$
Nothing wrong with the other answers. I just want to try my hand at this. I will prove by induction on the number of primes $n$ that
- For $K_n=\Bbb{Q}(\sqrt{p_1},\sqrt{p_2},\ldots,\sqrt{p_n})$ we have $[K_n:\Bbb{Q}]=2^n$ AND
- The extension $K/\Bbb{Q}$ is Galois with Galois group isomorphic to the $n$-fold Cartesian product $(C_2)^n$ generated by the automorphisms $\tau_i, i=1,2,\ldots,n,$ such that $\tau_i(\sqrt{p_j})=(-1)^{\delta_{ij}}\sqrt{p_j}$.
The base case $n=1$ is easy. Skipping that. Assume that the claim holds when we have $k$ primes.
Claim 1. $\sqrt{p_{k+1}}\notin K_k$.
Proof. Assume contrariwise that $\sqrt{p_{k+1}}\in K_k$. We know that $\Bbb{Q}(\sqrt{p_{k+1}})$ is a quadratic extension of $\Bbb{Q}$. By Galois theory the quadratic subfields of $K_k$ are exactly the fixed fields of index two subgroups of $\operatorname{Gal}(K_k/\Bbb{Q})$. By part two of the induction hypothesis, this Galois group is isomorphic to the additive group of a $k$-dimensional vector space $V$ over $\Bbb{F}_2$. The non-degenerate bilinear form, $B:V\times V\to\Bbb{F}_2, B\big((x_1,x_2,\ldots,x_k),(y_1,y_2,\ldots,y_k)\big)=\sum_{i=1}^kx_iy_i$, shows that the maximal subgroups are exactly the duals of the minimal subgroups of $V$. There are $2^k-1$ of those, namely the ones containing a single non-zero vector of $V$. Therefore $K_k$ has exactly $2^k-1$ quadratic subfields. But if $S$ is the product of any non-empty subset of $\{p_1,p_2,\ldots,p_k\}$, then $\Bbb{Q}(\sqrt{S})$ is a quadratic subfield of $K_k$. Such quadratic fields are easily seen to be distinct (only need the analogue of the argument showing $\sqrt2\notin\Bbb{Q}(\sqrt3)$ for this). Similarly we see that $\sqrt{p_{k+1}}$ is not in any of those quadratic subfields. The claim follows.
Claim 2. The inductive step is valid.
Proof. By Claim 1. $[K_{k+1}:K_k]=2$, so part 1 one of the induction hypothesis implies part 1 of the inductive step. Because $K_{k+1}$ is the splitting field of $\prod_{i=1}^{k+1}(x^2-p_i)\in\Bbb{Q}[x]$, it follows that $K_{k+1}$ is Galois over $\Bbb{Q}.$ Any automorphism $\tau\in \operatorname{Gal}(K_{k+1}/\Bbb{Q})$ is fully determined if we know the images $\tau(\sqrt{p_i}), i=1,2,\ldots,k+1$. There are two choices for each of those images (up to sign) - a total of $2^{k+1}$ combinations. Because the extension is Galois, we know that there will be exactly $2^{k+1}$ automorphisms, so all those sign combinations must occur. Q.E.D.
This is a minor variant of Jyrki's answer.
We want to prove the statements:
(1) If $p_1,...,p_n$ are distinct primes, then there are automorphisms $\tau_i$, ($i=1,...,n$), of $\mathbb Q[\sqrt{p_1},...,\sqrt{p_n}]$ such that $\tau_i\,\sqrt{p_j}=(-1)^{\delta_{ij}}\sqrt{p_j}$ for all $i,j$.
(2) If $p_1,...,p_n$ are distinct primes, then $\sqrt{p_n}\notin\mathbb Q[\sqrt{p_1},...,\sqrt{p_{n-1}}]$.
Clearly (1) and (2) are equivalent. [By this we mean that (1) holds for all $n$-tuples $(p_1,...,p_n)$ of distinct primes if and only if (2) holds for all such $n$-tuples.]
Assume that (2) is false, and let $(p_1,...,p_n)$ be a counterexample to (2) with $n$ minimum. In particular we have $$ \sqrt{p_n}\in K:=\mathbb Q[\sqrt{p_1},...,\sqrt{p_{n-1}}]. $$ Furthermore, the automorphisms $\tau_i$, ($i=1,...,n-1$), of $K$ are well-defined.
We easily check (3) and (4) below:
(3) For all $x$ in $K$ we have $\tau_i\,x=x$ if and only if $x$ is in the subfield generated by the $\sqrt{p_j}$ for $j\neq i$, and
(4) $\tau_i\,x=x$ for all $i$ if and only if $x$ is in $\mathbb Q$.
(5) We have $\tau_i\,\sqrt{p_n}=-\sqrt{p_n}$. (Indeed, in view of (3), the equality $\tau_i\,\sqrt{p_n}=\sqrt{p_n}$ would contradict the minimality of $n$.)
Now (5) and (4) imply that $$ \sqrt{\frac{p_n}{p_1\cdots p_{n-1}}} $$ is in $\mathbb Q$, which is easily seen to be false.