I guess you could argue by induction as follows. Let
$$
E = \mathbb Q(\sqrt {2},\sqrt {3},\sqrt {5},\dotsc,\sqrt {p_{n-1}}) ,
$$
and suppose that $E/ \mathbb Q$ has degree $2^{n-1}$. Assume also by induction that the Galois group $G$ of the Galois extension $E/\mathbb Q$ is elementary abelian, sending $\sqrt{p_{i}} \mapsto \pm \sqrt{p_{i}}$. If, by way of contradiction, $\sqrt{p_{n}} \in E$, then
$$
\mathbb Q(\sqrt{p_{n}}) = \mathbb Q(\sqrt{p_{i_1} \dots p_{i_k}}),
$$
for some $1 \le i_1 < \dots < i_k < n$. (I am appealing to the Galois correspondence here.) So
$$
a + b \sqrt{p_{n}} = \sqrt{p_{i_1} \dots p_{i_k}},
$$
for some $a, b \in \mathbb Q$. Square both sides and distinguish a couple of cases to get a contradiction.
PS Let me expand a bit on the Galois connection here. $G$ is generated by the $n-1$ elements $\sigma_i$, that fix all $\sqrt{p_{j}}$, for $j \ne i$, and send $\sqrt{p_{i}} \mapsto - \sqrt{p_{i}}$. Intermediate fields $F$ of $E/\mathbb{Q}$ such that $F/\mathbb{Q}$ has degree two correspond to maximal subgroups of $G$. Each such maximal subgroup $M$ is defined by an equation
$$
a_{i_1} + \dots + a_{i_k} = 0,
$$
in the exponents of an element
$$
\sigma_1^{a_1} \cdot \dots \cdot \sigma_{n-1}^{a_{n-1}} \in G
$$
where you may think $a_i \in \mathbb{Z}_2$. (An elementary abelian group of order a power of 2 is really a vector space over $\mathbb{Z}_2$.) Now it is easy to verify that the elements of $M$ fix $\sqrt{p_{i_1} \dots p_{i_k}}$.
