Proving $I\subseteq\sqrt I$ in a commutative ring

Question

Let $R$ be a commutative ring and $I$ be an ideal of $R$. How can this directly be proved?

P.S: I tried to take an element from the left side and then get to the other side but it didn't work out. I don't know what I'm missing.

Answer 1

Note that $\sqrt{I} = \{a \in R : a^{n} \in I$ for some $n \in \mathbb{N} \}$

Take $a \in I \subset R$ then $a^{1} = a \in I$, then $a \in \sqrt{I}$