<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] Subtract the first row from all of the other rows ... we have
\begin{eqnarray*}
\left[\begin{array}{cccccc}
-x  & 1      & 1 & 1 &\cdots &1\\
1+x & -(x+1) & 0 & 0 &\cdots &0\\ 
1+x & 0 &-(x+1)  & 0 &\cdots &0\\ 
1+x & 0 &0 & -(x+1)  & \cdots &0\\ 
\vdots & \vdots &\vdots &\vdots &\ddots &\vdots & \\
1+x & 0 &0 & 0  & \cdots &-(x+1)\\ 
\end{array}\right]
\end{eqnarray*}
The leading diagonal will give a contribution of $\color{blue}{-x(-1)^{n-1}(x+1)^{n-1}}$ and if we pick the $i^{th}$ $1$ in the first row then we must pick the $i^{th}$ $(1+x)$ in the first column and the rest of the terms on the leading diagonal, there are $\color{red}{(n-1)}$ of these & each will give a contribution of $\color{orange}{-(1+x)^{n-1}(-1)^{n-2}}$. Putting these together we have $(-1)^{n}(1+x-n)(1+x)^{n-1}$.
\begin{eqnarray*}
\color{blue}{-x(-1)^{n-1}(x+1)^{n-1}}+\color{red}{(n-1)} \color{orange}{(-(1+x)^{n-1}(-1)^{n-2})}=(-1)^{n}(1+x-n)(1+x)^{n-1}.
\end{eqnarray*}
</s>