I want to prove $\mathbb{Q}(\sqrt{p_1},\dots,\sqrt{p_{n-1}})=\mathbb{Q}(\sum\limits_{i =1}^n\sqrt{p_i})$, where $p_i$ are different prime integers.
But I must solve this problem firstly: $\sqrt{p_n} \notin \mathbb{Q}(\sqrt{p_1},\dots,\sqrt{p_{n-1}})$.
It's trivial when $n=2$, but I don't have any idea for $n>2$.