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 $[F_n:\mathbb Q]=p_n\cdot P_{n-1}=P_n$