Conjugate transpose arithmetic question

Question

There's this really simple question that's been bugging me since I can't seem to finish it. Let $A$ be an $n\times n$ matrix with complex entries, and $A^*$ its conjugate transpose. Given $A^*=A^7$, show $A^8=I$.

I tried using the property $(M^*)^*=M$. That gives me $A=(A^7)^*=(A^*)^7=(A^7)^7=A^{49}$. I think this implies $I=A^{48}=(A^8)^6$. So there's an $A^8$ and an $I$ but I don't know what to do about the $6^{th}$ power (or if I'm even doing this right).

Answer 1

As symplectomorphic pointed out, the zero matrix is a simple counterexample, thus it cannot be shown that $A^8=I$.

Answer 2

For real $A$, if $A$ is an orthogonal matrix, $A^{\intercal}A=I$ and therefore $A^{\intercal}=A^{-1}$.

Now since $A^{\intercal}=A^{7}$, then $A^{8}=AA^{7}=AA^{\intercal}=AA^{-1}=I$.

More generally for complex $A$, if $A$ is unitary ($A^*=A^{-1}$) the same result can be achieved.