Let $A$ be an $n\times n$ invertible complex matrix such that $A^7 = A^*$. Show that $A^8 = I$. 
Let $A$ be an $n\times n$ invertible complex matrix such that $A^7 = A^*$ (where $*$ denotes conjugate transpose). Show that $A^8 = I$.

Here are my thoughts so far:


*

*I was able to show that all the eigenvalues of $A$ satisfy $\lambda^8 = 1$.

*I tried writing $A$ in Jordan canonical form $A = PJP^{-1}$ so that $A^* = (P^{-1})^*J^*P^*$ and $A^7 = PJ^7P^{-1}$. I was hoping to conclude $J^* = J^7$ but this isn't necessarily true as they may not be Jordan matrices; they would be if $J$ was diagonal, but if I knew that $J$ was diagonal, I would be done.

*As $A^{49} = (A^7)^7 = (A^*)^7 = (A^7)^* = (A^*)^* = A$, $A^{48} = I$ so the minimal polynomial of $A$ divides $x^{48}-1$; note that $x^8 - 1$ is a factor of $x^{48}-1$.

*If $A^8 = PJP^{-1}$ is the Jordan normal form of $A^8$, then $J = I + N$ and $J^6 = I$ by the previous point so $(I+N)^6 = I$ but I can't directly deduce from this equality that $N = 0$.
Any hints are very much appreciated.
 A: In regards to your last point, it suffices to work on one Jordan block at a time. Let $N$ be the nilpotent part of an $e\times e$ Jordan block. Examine $\{I,N,N^2,\cdots,N^{e-1}\}$, see what the elements look like explicitly. In particular, consider linear dependence or independence.
A: I use $A^\dagger$ for the Hermitian adjoint, or conjugate transpose, of $A$.
We have
$A^7 = A^\dagger; \tag{1}$
this implies that $A$ is normal, or $AA^\dagger = A^\dagger A$:
$AA^\dagger = AA^7 = A^7A = A^\dagger A; \tag{2}$
since $A$ is normal, there is a unitary matrix $U$ which diagonalizes $A$, thus:
$U^\dagger A U = \Lambda, \tag{3}$
where $\Lambda$ is a diagonal matrix whose diagonal is comprised of the eigenvalues of $A$; see this widipedia entry on normal matrices.  Furthermore,
$U^\dagger A^\dagger U = (U^\dagger A U)^\dagger = \Lambda^\dagger, \tag{4}$
and $\Lambda$ being diagonal, we have $\Lambda^\dagger = \Lambda^\ast$, the matrix of complex conjugates of $\Lambda$; this illustrates the reason I chose ${}^\dagger$ for adjoint; I needed to reserve ${}^\ast$ for (elementwise) conjugation.  In any event, (4) thus yields
$U^\dagger A^\dagger U = \Lambda^\dagger = \Lambda^\ast. \tag{5}$
Now (1) implies
$(U^\dagger A U)^7 = U^\dagger A^7 U = U^\dagger A^\dagger U, \tag{6}$
so by (3) and (4)
$\Lambda^7 = \Lambda^\ast; \tag{7}$
this shows that for any diagonal entry $\lambda$ of $\Lambda$ we must have
$\lambda^7 = \lambda^\ast, \tag{8}$
thus
$\vert \lambda \vert^7 = \vert \lambda^7 \vert = \vert \lambda^\ast \vert = \vert \lambda \vert; \tag{9}$
and since $A$ is invertible, $\lambda \ne 0$, so
$\vert \lambda \vert^6 = 1, \tag{10}$
whence we must have
$\vert \lambda \vert = 1, \tag{11}$
$\vert \lambda \vert$ being positive real.  All the $\lambda$ are therefore unimodular, yielding
$\lambda^\ast = \lambda^{-1}; \tag{12}$
Applying (12) to (8) gives
$\lambda^8 = 1, \tag{13}$
hence
$\Lambda^8 = I. \tag{14}$
Finally, (3) implies
$A^8 = (U \Lambda U^\dagger)^8 = U I U^\dagger = I, \tag{15}$
as per request.  QED.
Nota Bene:  There seems to be no magic in the hypothesis $A^7 = A^\dagger$, other than the general magic of mathematics:  apparently $A^n = A^\dagger$ implies $A^{n + 1} = I$ for any integer $n \ge 2$ and invertible $A$.  End: Nota Bene.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
A: Since $A^7=A^*$ we have $A^*A=AA^*$ so $A$ is a normal matrix. Thus $A$ is diagonalisable, and since all its eigenvalues are roots of $X^8-1=0$ we conclude that $A^8=I$.
Edit: Indeed, consider $x\ne0$ an eigenvector of $A$ corresponding to the eigenvalue $\lambda$.
We have
$$
\bar{\lambda}\Vert x\Vert^2=\langle x,Ax\rangle=\langle A^*x,x\rangle
=\langle A^7x,x\rangle=\lambda^7\Vert x\Vert^2
$$
Thus $\lambda^7=\bar{\lambda}$. In particular, $|\lambda|^7=|\lambda|$, and since $\lambda\ne0$ because $A$ is invertible we conclude that $|\lambda|=1$, and consequently
$\lambda^7=\bar{\lambda}=1/\lambda$ or $\lambda^8=1$.
