Ramification in $\mathbb{Q}$ adjoint square roots of prime numbers

Let $K = \mathbb{Q}(\sqrt{p_1}, ..., \sqrt{p_n})$, where the $p_i$'s are distinct prime numbers. Let $p$ be a prime such that $p \neq p_i$ for all $i$, and $p \neq 2$. Why is it true that $p$ doesn't ramify in $K$?

Let $F$ be a number field, and let $L_1, L_2$ be two extensions of $F$. Let $L = L_1\cdot L_2$.
A prime $\mathfrak p$ of $F$ is unramified in $L$ if and only if it is unramified in both $L_1$ and $L_2$.
In your setting, since $p$ is unramified in $\mathbb Q(\sqrt{p_i})$ for each $i$, it is unramifed in the compositum of these fields, which is $K$.