Calculation of determinant of an arrowhead matrix

Is there any easier way to make sure the determinant of the following $n \times n$ matrix is $n$?

$$\begin{vmatrix} 1 & -1 & -1 & -1 & \cdots & -1 \\ 1 & 1 & 0 & 0 & \cdots & 0 \\ 1 & 0 & 1 & 0 & \cdots & 0 \\ 1 & 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & 0 & 0 & 0 &\cdots & 1 \end{vmatrix} = n$$

I figured it with a smaller dimension and it indeed produces the determinant that is the size of dimension. I tried to do a cofactor expansion with the first row, and each term produces the determinant of $1$ and if you sum them up, then the total determinant will be $n$. But the sign change for each cofactor is confusing, and it is not easily seen that each cofactor term is actually positive $1$.

$$ \begin{vmatrix} 1 & -1 & -1 & -1 & \cdots & -1 \\ 1 & 1 & 0 & 0 & \cdots & 0 \\ 1 & 0 & 1 & 0 & \cdots & 0 \\ 1 & 0 & 0 & 1 & & 0 \\ \vdots & \vdots & \vdots & & \ddots & \vdots \\ 1 & 0 & 0 & \cdots & 0 & 1 \end{vmatrix} = \begin{vmatrix} n & 0 & 0 & 0 & \cdots & 0 \\ 1 & 1 & 0 & 0 & \cdots & 0 \\ 1 & 0 & 1 & 0 & \cdots & 0 \\ 1 & 0 & 0 & 1 & & 0 \\ \vdots & \vdots & \vdots & & \ddots & \vdots \\ 1 & 0 & 0 & \cdots & 0 & 1 \end{vmatrix} = n $$

Note that in a triangular matrix its determinant equals the product of the diagonal entries.