Edit: Crap, even my hypothesis was wrong. If you put $A = \left[ \begin{array}{cc} 1&-1\\3&-2 \end{array} \right]$, then $A^3 = I$ but no eigenvalue is $1$. (What's true is that all eigenvalues are $n$th roots of unity.)
I ask this because I believe it might resolve a trickier problem I'm working on in a simple way. Obviously this is false if $n$ is even, but I have no counterexample if $n$ is odd. Indeed, if $x$ is an eigenvector of $A$,
$$A^nx = \lambda^nx = Ix = x \implies \lambda = 1 $$
so $\lambda = 1$ is the only eigenvalue of the matrix. This says $\operatorname{trace}(A) = n$ and $p(\lambda) = (1 - \lambda)^n$. (Trivially, $\det(A) = 1$ from the multiplicativity of the determinant.)
(Incidentally, I am aware that there is a subtlety here in assuming a real eigenvector exists, since the rotation-by-$90^{\circ}$ matrix satisfies $A^4 = I$ and has none; but it seems that the condition that $A^n = I$ for odd $n$ may take care of this problem.)
How can we prove this, in a fairly simple way? (If the proof is complicated, I'm still game, but it's probably not the intended solution to my original problem.)
It is not true. A simple counterexample for $n=3$ is $$A=\left(\begin{array}{ccc} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right).$$ More complicated counterexamples for any $n$ can be constructed using permutation matrices.