Let $n$ be a natural integer with $n \geqslant 2$. For any real number $x$, we consider the following matrix in $\mathscr{M}_n(\mathbb{R})$
$$M_x = \left(\begin{array}{ccccc} x & 1 & \cdots & 1 & 1 \\ 1 & x & \cdots & 1 & 1 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & 1 & \cdots & x & 1 \\ 1 & 1 & \cdots & 1 & x \end{array}\right).$$
Deduce that for all $x \in \mathbb{R}$, we have
$$\sum_{\sigma \in \mathfrak{S}_n} \varepsilon(\sigma) x^{\nu(\sigma)} = (x-1)^{n-1}(x+n-1).$$