If $A$ is context-free then $A^*$ is regular
Let $A=\{a^nb^n \mid n \in \mathbb{N}\}$. Then we know that $A$ is non-regular and context-free. Also we can see that $A^*\cap a^*b^*=A$. Since $a^*b^*$ is a regular expression, we do know that it is regular. Lets assume that A* is regular.
The regular languagues are closed unter intersection. Therefore $A^*\cap a^*b^*$ must be also regular(because we assume that $A^*$ is regular). This would implicate that A is regular because  $A^*\cap a^*b^*=A$. This is a contradiction because we know that A is not regular. Therefore A* cant be regular.
$q.e.d$