<s>[INST] On a determinant identity 
Show that
$$\det \begin{bmatrix} -x & 1 & \dots & 1 \\ 1 & -x & \dots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \dots & -x \end{bmatrix}  = (-1)^n(1+x-n)(1+x)^{n-1}$$
for all $n \in \mathbb{N}$.

I tried using induction but I didn't manage (using row/column swaps). Does anyone have any idea?
 [/INST] HINT
$$\begin{align}\begin{vmatrix} -x & 1 & \dots & 1 \\ 1 & -x & \dots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \dots & -x \end{vmatrix}&=\begin{vmatrix} -x+n-1& 1 & \dots & 1 \\ -x+n-1 & -x & \dots & 1 \\ \vdots & \vdots & \ddots & \vdots \\-x+n-1 & 1 & \dots & -x \end{vmatrix}\\&=(-x+n-1)\begin{vmatrix} 1& 1 & \dots & 1 \\ 1 & -x & \dots & 1 \\ \vdots & \vdots & \ddots & \vdots \\1 & 1 & \dots & -x \end{vmatrix}\\&=(-x+n-1)\begin{vmatrix} 1& 1 & \dots & 1 \\ 0 & -x-1 & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\0 & 0 & \dots & -x-1 \end{vmatrix}\end{align}$$
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