Algebraic Field Extension of recursive subfields 
Let $p_n$ be the n$^{th}$ prime and define subfields $F_n$ of $\mathbb{R}$ recursively by $F_n$=$F_{n-1}(\sqrt[p_n]{p_n})$. Show that
  $$[F_n:\mathbb{Q} ]=\prod_{i=1}^n p_i$$ and deduce that an algebraic extension need not be finite.

Any help would be greatly appreciated. Best regards.
 A: Check that $[\mathbb{Q}(\sqrt[p_n]{p_n}):\mathbb{Q}]=p_n$. Suppose by induction $[F_{n-1}:\mathbb{Q}]=\prod_{k=1}^{n-1}p_k$. Notice $[\mathbb{Q}(\sqrt[p_n]{p_n}):\mathbb{Q}] \mid [F_n:\mathbb{Q}]$ and that $[\mathbb{Q}(\sqrt[p_n]{p_n}):\mathbb{Q}]\geq [F_n:F_{n-1}]$.
A: Let's do it by induction and let me put $P_n=\prod_{i=1}^n p_i$
The statement  $[F_n:\mathbb Q]=P_n$ is clear for $n=1$.
Assume it is true for for $n-1$ and let's prove it for $n$.
Let $K=\mathbb Q(\sqrt[p_n](p_n))$.
It is clear that $[F_n:F_{n-1}]\leq p_n$ since $X^{p_n}-p_n$ kills $\sqrt[p_n](p_n)$ and has coefficients in $F_{n-1}$. Hence
$$[F_n:\mathbb Q]= [F_n:F_{n-1}][F_{n-1}:\mathbb Q] \leq p_n\cdot P_{n-1}$$ 
On the other hand we have $F_{n-1}\subset F_n$ implying that $P_{n-1}|[F_n:\mathbb Q]$ and also $K\subset F_n$ implying $p_n|[F_n:\mathbb Q]$.
Since $P_{n-1}$ and $p_n$ are relatively prime (this is the key point !) we deduce 
$$ p_n\cdot  P_{n-1}|[F_n:\mathbb Q]           $$
The two displayed equations prove that $[F_n:\mathbb Q]=p_n\cdot P_{n-1}=P_n$
