$\sqrt{\sqrt{\mathfrak{a}}} = \sqrt{\mathfrak{a}}$
$\sqrt{\sqrt{\mathfrak{a}}} \subseteq \sqrt{\mathfrak{a}}:$
Let $x \in \sqrt{\sqrt{\mathfrak{a}}}$, then $x^n \in \sqrt{\mathfrak{a}}$ for some $n > 0$. Consider some $m > 0$ such that $x^{nm} = (x^n)^m \in \mathfrak{a}$. Therefore, $x \in \sqrt{\mathfrak{a}}$ and $\sqrt{\sqrt{\mathfrak{a}}} \subseteq \sqrt{\mathfrak{a}}$.
$\sqrt{\mathfrak{a}} \subseteq \sqrt{\sqrt{\mathfrak{a}}}:$
Let $x \in \mathfrak{a}$. Then $x^1 = x \in \mathfrak{a}$, thus $x \in \sqrt{\mathfrak{a}}$, i.e., $\mathfrak{a} \subseteq \sqrt{\mathfrak{a}}$. So, consider $x \in \sqrt{\mathfrak{a}}$. Then $\exists n>0$ such that $x^n \in \mathfrak{a} \subseteq \sqrt{\mathfrak{a}}$. Thus $x^n \in \sqrt{\mathfrak{a}}$ for some $n>0$. Therefore $x \in \sqrt{\sqrt{\mathfrak{a}}}$. Hence $\sqrt{\mathfrak{a}} \subseteq \sqrt{\sqrt{\mathfrak{a}}}$.
Is this proof correct? Any help in correcting any stupid mistakes would be greatly appreciated. Thank you for the help!!!