$n^{th}$ determinant Find determinant $D_n$ of matrix $$
        \begin{bmatrix}
        1 & 1 & \cdots & 1 & -n \\
        1 & 1 & \cdots & -n & 1 \\
        \vdots & \vdots & \ddots & \vdots & \vdots \\
        1 & -n & \cdots & 1 & 1 \\
        -n & 1 & \cdots & 1 & 1
        \end{bmatrix}
$$
After multiplying first row by $-1$ and adding to $n-1$ rows:
$$
        \begin{vmatrix}
        1 & 1 & \cdots & 1 & -n \\
        0 & 0 & \cdots & -n-1 & n+1 \\
        \vdots & \vdots & \ddots & \vdots & \vdots \\
        0 & -n-1 & \cdots & 0 & n+1 \\
        -n & 1 & \cdots & 1 & 1
        \end{vmatrix}
$$
Second and $n-1$ row are changing place:
$$
        \begin{vmatrix}
        1 & 1 & \cdots & 1 & -n \\
        0 & -n-1 & \cdots & 0 & n+1 \\
        \vdots & \vdots & \ddots & \vdots & \vdots \\
        0 & 0 & \cdots & -n-1 & n+1 \\
        -n & 1 & \cdots & 1 & 1
        \end{vmatrix}
$$
Multiplying first row by $n$ and adding it to $n$-th row:
$$
        \begin{vmatrix}
        1 & 1 & \cdots & 1 & -n \\
        0 & -n-1 & \cdots & 0 & n+1 \\
        \vdots & \vdots & \ddots & \vdots & \vdots \\
        0 & 0 & \cdots & -n-1 & n+1 \\
        0 & n+1 & \cdots & n+1 & 1-n^2
        \end{vmatrix}
$$
Adding second to $n$-th row:
$$
        \begin{vmatrix}
        1 & 1 & \cdots & 1 & -n \\
        0 & -n-1 & \cdots & 0 & n+1 \\
        \vdots & \vdots & \ddots & \vdots & \vdots \\
        0 & 0 & \cdots & -n-1 & n+1 \\
        0 & 0 & \cdots & n+1 & (1+n)(2-n)
        \end{vmatrix}
$$
Adding $n-1$ to $n$-th row:
$$
        \begin{vmatrix}
        1 & 1 & \cdots & 1 & -n \\
        0 & -n-1 & \cdots & 0 & n+1 \\
        \vdots & \vdots & \ddots & \vdots & \vdots \\
        0 & 0 & \cdots & -n-1 & n+1 \\
        0 & 0 & \cdots & 0 & (1+n)(3-n)
        \end{vmatrix}
$$
Product on main diagonal gives $$D_n=(n+1)(3-n)(-n-1)^{n-2}$$
This is not correct. What is wrong with this upper triangular transformation?

Another solution, perhaps with an unusual method.
Let $T_m(x)$ the determinant of the $m\times m$ matrice, with entries $x$ on the antidiagonal, and $1$ otherwise. Clearly, your determinant $D_n$ is equal to $T_n(-n)$. The determinant $T_m(x)$ is a polynomial in $x$. The derivative of $T_m$ is obtained by differentiating the $k$-th column, computing the new determinant, and adding all these determinants. The computation of these determinant is easy: they are all equal to $(-1)^{m-1}T_{m-1}(x)$. Hence we get $T_m^{\prime}(x)=(-1)^{m-1}m T_{m-1}(x)$. In addition it is clear that $T_m(1)=0$. Put $S_m(x)=T_m(1+x)$. We get $S_m^{\prime}(x)=(-1)^{m-1}m S_{m-1}(x)$ and $S_m(0)=0$. We have $S_1(x)=1+x$, $S_2(x)=-2x-x^2$. An easy induction show that $S_m(x)=(-1)^{m(m-1)/2}(m x^{m-1}+x^m)$. Now to finish
$D_n=T_n(-n)=S_n(-n-1)$, and we are done.
