If A is context free then A* is regular


I am currently studying for my exam and I am having trouble to solve this question :

If A is context free then A* is regular . (Wrong or right)

I think its wrong because if A is context free it means that A can be a non regular languague. And the non regular languagues are not closed under the kleene star operation(At least I think so). I am not sure how write this in a more formal way.

Maybe like this ? : Let A=$ \{a^nb^n|n \in \mathbb{N}\}$ . Then we know that A is non regular and context free. Then A*= I am not sure about this :/