Does $\mathbb{Q}(\sqrt{m_1},...,\sqrt{m_k})$ has degree $2^k$ over $\mathbb{Q}?
Elementary proof for $\sqrt{p_{n+1}} \notin \mathbb{Q}(\sqrt{p_1}, \sqrt{p_2}, \ldots, \sqrt{p_n})$ where $p_i$ are different prime numbers.