Let $e_k$ be the $k$-th vector of the canonical base of $\mathbb R^n$ and let $$B = [e_2 \mid e_3 \mid \dots \mid e_n \mid e_1]$$
What it the quickest way to show that the charachteristic polynomial of $B$ is $\lambda^n - 1$?
My work
I can show that if $\lambda$ is an eigenvalue of $B$, then $\lambda^n = 1$ (trivially because $B^n = I$)but this does not suffice as those $\lambda$ can all be equal to one; I want to show that they are distinct.
As I did not know how to do that, I resorted to use the fact (which is obvious, altough it should be proved more formally) that $tr(B^k) = 0$ for all $k < n$. Then, as the coefficients in the characteristic polynomial of $B$ (except the first and last one) are all polynomials in the variables $tr(B), tr(B^2), \dots , tr(B^{n-1})$ they are all equal to $0$. And since $\det(B) = (-1)^{n+1}$, we have $$P_B(\lambda) = \lambda^n + (-1)^n(-1)^{n+1} = \lambda^n - 1$$
But it seems like an overkill, especially because I do not know how to prove that those coefficients are indeed polynomials in $tr(B^i)$. Can someone suggest a more elegant way?
Observe that $I, B, B^2, \ldots, B^{n-1}$ are linearly independent (because their nonzero entries lie on entirely different diagonals). Hence no nonzero degree-$(n-1)$ polynomial over $\mathbb R$ can annilhilate $B$. In other words, $x^n-1$ is the minimal polynomial. You may continue from here.
Edit. On second thought, perhaps the quickest way to prove the assertion is to calculate $\det(xI−B)$ using Laplace expansion along the first row: $$ (-1)^{1+1}x\det\pmatrix{x\\ -1&\ddots\\ &\ddots&\ddots\\ &&\ddots&\ddots\\ &&&-1&x} +(-1)^{1+n}(-1)\det\pmatrix{-1&x\\ &-1&\ddots\\ &&\ddots&\ddots\\ &&&\ddots&x\\ &&&&-1}. $$