Showing field extension $\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5})/\mathbb{Q}$ degree 8 [duplicate]

An easy way to show that $\sqrt{5}$ is not in $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is to note that $[\mathbb{Q}(\sqrt{2},\sqrt{3})\colon\mathbb{Q}]=4$, so by the Fundamental Theorem of Galois Theory it has either one or three intermediate fields of degree $2$ over $\mathbb{Q}$ (since a group of order $4$ has either a single subgroup of index $2$, when the group is cyclic, or exactly three, when it is the Klein $4$-group). Since this field contains the three distinct intermediate field $\mathbb{Q}(\sqrt{2})$, $\mathbb{Q}(\sqrt{3})$, and $\mathbb{Q}(\sqrt{6})$, it cannot also contain the field $\mathbb{Q}(\sqrt{5})$, which is distinct from those three.

For your second question, the answer is likewise "yes". The only intermediate fields of degree $2$ are $\mathbb{Q}(\sqrt{d})$, where $d$ is the of some (but at least one) of the $p_i$. You can prove it by induction on $n$.