# [Some punny title about determinants and fixed points]

Prove the following identity

$\sum_\sigma (-1)^\sigma t^{\mathrm{Fix}\;\sigma} = \det\begin{pmatrix} t & 1 &\cdots& 1& 1\\ 1 & t &\cdots& 1& 1 \\ \vdots & \vdots &\ddots & \vdots & \vdots \\ 1 & 1 &\cdots& t& 1\\ 1 & 1 &\cdots& 1& t \end{pmatrix}$

the sum running through $S_n$ and the matrix being of size $n\times n$.

1. It follows directly from the Leibniz’s formula for determinants: $det(a_{ij}) = \sum_{\sigma \in S_n}(-1)^{\sigma}\prod_{i=1}^na_{\sigma(i)i}$