Permutation matrix/Cycle/Characteristic polynomial/Fact/Proof

We may assume that the cycle has the form ${\displaystyle {}1\mapsto 2\mapsto 3\mapsto \ldots \mapsto k\mapsto 1}$. The corresponding permutation matrix ${\displaystyle {}M_{\rho }}$ looks with respect to ${\displaystyle {}e_{k+1},\ldots ,e_{n}}$ like the identity matrix and has, with respect to the first ${\displaystyle {}k}$ standard vectors, the form

${\displaystyle {\begin{pmatrix}0&0&\ldots &0&1\\1&0&0&\ldots &0\\0&1&0&\ldots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\ldots &0&1&0\end{pmatrix}}.}$

The determinant of ${\displaystyle {}XE_{n}-M_{\rho }}$ is ${\displaystyle {}(X-1)^{n-k}}$ multiplied with the determinant of

${\displaystyle {\begin{pmatrix}X&0&\ldots &0&-1\\-1&X&0&\ldots &0\\0&-1&X&\ldots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\ldots &0&-1&X\end{pmatrix}}.}$

The expansion with respect to the first row yields

${\displaystyle {}X^{k}+(-1)^{k+1}(-1)(-1)^{k-1}=X^{k}-1\,.}$