If $A$ is normal and $A$ and $B$ commute, then $A^*$ and $B$ commute Let $A$ is a normal matrix: $A^*\! A = A A^*\!\!$,$\,$ and $AB = BA$. Prove that $A^*\!B=BA^*\!\!$.

I can prove that if $\det A\ne 0$ by multiplication $AB=BA$ by $A^*$ left and right and using some manipulation. But I have no idea what to do if $\det A = 0$.
Hint: If $A$ is normal then $A^*$ is a polynomial in $A$. 

$\bf{Added:}$

Since $A$ is normal there exists $U$ a unitary matrix so that $A = U D U^*$ ( $A$ is unitarily diagonalizable), and so $A^* = U \bar D U^*$. Let $\lambda_k$ be the eigenvalues of $A$ ( the diagonal of $D$). Take $P$ a polynomial with complex coefficients so that $$P(\lambda_k) = \bar \lambda_k$$ Check that $P(A) = A^*$
In fact, if $|A|\neq 0$, then we can find a sequence $A_n$ such that $\lim_{n\rightarrow \infty}A_n=A$ and every term $A_n$ is inverse, that is, $|A_n|\neq 0$.