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?

$$\begin{bmatrix} -x & 1 & \dots & 1 \\ 1 & -x & \dots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \dots & -x \end{bmatrix} = 1_n 1_n^\top - (x+1) \mathrm I_n = - (x+1) \left( \mathrm I_n - \frac{1}{x+1} 1_n 1_n^\top \right)$$
Using the Weinstein-Aronszajn determinant identity,
$$\begin{array}{rl} \det \begin{bmatrix} -x & 1 & \dots & 1 \\ 1 & -x & \dots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \dots & -x \end{bmatrix} &= (-1)^n (x+1)^n \det \left( \mathrm I_n - \frac{1}{x+1} 1_n 1_n^\top \right)\\ &= (-1)^n (x+1)^n \left( 1 - \frac{1}{x+1} 1_n^\top 1_n \right)\\\\ &= (-1)^n (x+1)^n \left( 1 - \frac{n}{x+1} \right)\\\\ &= (-1)^n (x+1)^{n-1} \left( x+1 - n \right)\end{array}$$
