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 $n>2$.
Generalise the assumption from primes to assuming that no product is a perfect square. The field $\mathbb{Q}(\sqrt{p_1},\ldots ,\sqrt{p_n}) $ has $2^n-1$ many subfields of order $2$ namely for $X\subseteq n$ and $X\neq \emptyset$, there is the subfield, $$\mathbb{Q}(\sqrt{\prod\limits_{i\in X}p_i}) $$ and these are distinct by the better assumption used in the $n=2$ case. If $\sqrt{p_n}\in \mathbb{Q}(\sqrt{p_1},\ldots ,\sqrt{p_{n-1}}) $ then the Galois group of $\mathbb{Q}(\sqrt{p_1},\ldots ,\sqrt{p_n}) $ would have size at most $2^{n-1}$ and thus there could not be enough subfields of degree $2$.