I know that this questions has asked before, but I didn't get the solution.
I'm studying Field extensions, so I've found the question below.
Denote $(p_n)_{n \in \mathbb{N}}$ the sequence of the first prime numbers. For instance $p_1=2, p_2=3,$etc. Now, denote $F_0 = \mathbb{Q}$ and $F_n = \mathbb{Q}(\sqrt{p_1}, \sqrt{p_2}, \dots, \sqrt{p_n})$. Show that $[F_n: F_{n-1}]=2, \forall n \in \mathbb{N}, n\geq 1$.
$\textbf{My attempt:}$
I have proved that $[F_1: F_0]=2$, because $\sqrt{2}$ is root of $x^2-2$, which is irreducible of $\mathbb{Q}$.
I have tried to prove by induction, that is, suppose that there is $k \in \mathbb{N}$ s.t $[F_k:F_{k-1}]=2$, but it doesn't help to prove that $[F_{k+1}:F_k]=2$.
I know that $[F_{k+1}:F_k] \leq 2$, because $\sqrt{p_{k+1}}$ is root of $x^2 - p_{k+1}$, and it is a polynomial in $\mathbb{Q}[x]$ which is a subset of $ F_k[x]$.
Futhermore, $[F_{k+1}:F_k]=1 \iff \sqrt{p_{k+1}} \in F_k$, but it is hard to prove that its not happens (because it's base is very big).
The last attempt was:
Suppose that $\sqrt{p_{k+1}} \in F_k$ then the element $\sqrt{p_1} + \sqrt{p_2}+ \cdots + \sqrt{p_n}+ \sqrt{p_{n+1}}$ is in $F_k$, and I've tried to prove that it is root of a irreducible polynomial in $\mathbb{Q}[x]$, with degree $2^{n+1}$ then $[F_{k+1}:F_0]= 2^{2k+1}$, but it didn't work to.
So, can you help me??
