Help me with the result of this determinant.. $$
D =
\begin{vmatrix}
1 & 1 & 1 & \dots & 1 & 1 \\ 
2 & 1 & 1 & \dots & 1 & 0 \\
3 & 1 & 1 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\
n-1 & 1 & 0 & \dots & 0 & 0 \\
n & 0 & 0 & \dots & 0 & 0 \\
\end{vmatrix}
=n*1*(-1)^\frac{n(n-1)}{2}
$$

I don't quite understand the solution of this determinant. I do understand that if we use Laplace expansion along the last row we get
$$
D = n*
\begin{vmatrix}
1 & 1 & 1 & \dots & 1 & 1 \\ 
1 & 1 & 1 & \dots & 1 & 0 \\
1 & 1 & 1 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\
1 & 1 & 0 & \dots & 0 & 0 \\
1 & 0 & 0 & \dots & 0 & 0 \\
\end{vmatrix}
$$
But how does the remaining determinant euqal: $1*(-1)^\frac{n(n-1)}{2}$?

Edit:

$$
\begin{vmatrix}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0 \\
\end{vmatrix}
=(-1)^{4}
\begin{vmatrix}
0 & 1 \\
1 & 0 \\
\end{vmatrix}
=(-1)^{4+3+2}
$$
I thought it should go: $(-1)^{3+2+1}$ or is the power actually the sum of row and column coordinates?
 A: To compute
$$\begin{vmatrix}
1 & 1 & 1 & \dots & 1 & 1 \\ 
1 & 1 & 1 & \dots & 1 & 0 \\
1 & 1 & 1 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\
1 & 1 & 0 & \dots & 0 & 0 \\
1 & 0 & 0 & \dots & 0 & 0 \\
\end{vmatrix}$$
You can proceed like that: If we swap the $k$-th row with the $n-k$-th row of the matrix for $k=1,2,\ldots$ then, at some point, we will get a lower triangular matrix with only $1$ on its main diagonal (so its determinant is $1$). Now, each swap corresponds to an elementary operation of the Gauss-Jordan method and thus changes the sign of the determinant (i.e. multiplies it by $-1$). So count the number of swaps to get the result.
A: First do an elementary transformation that does not change the determinant, namely add to the first column the sum of all the others. You are left with
$$
\begin{vmatrix}
n & 1 & 1 & \dots & 1 & 1 \\ 
n & 1 & 1 & \dots & 1 & 0 \\
n & 1 & 1 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\
n & 1 & 0 & \dots & 0 & 0 \\
n & 0 & 0 & \dots & 0 & 0 \\
\end{vmatrix}
=
n \cdot \begin{vmatrix}
1 & 1 & 1 & \dots & 1 & 1 \\ 
1 & 1 & 1 & \dots & 1 & 0 \\
1 & 1 & 1 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\
1 & 1 & 0 & \dots & 0 & 0 \\
1 & 0 & 0 & \dots & 0 & 0 \\
\end{vmatrix}
=
n \cdot (-1)^{s}.
$$
Here $s$ is the sign of the permutation $\sigma$ on $\{1, 2, \dots, n\}$ that swaps $i$ with $n-i$, since exchanging the $i$-th row with the $(n-i)$-th row of
$$
\begin{bmatrix}
1 & 1 & 1 & \dots & 1 & 1 \\ 
1 & 1 & 1 & \dots & 1 & 0 \\
1 & 1 & 1 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\
1 & 1 & 0 & \dots & 0 & 0 \\
1 & 0 & 0 & \dots & 0 & 0 \\
\end{bmatrix},
$$
for $1 \le i \le n/2$, we get a matrix which has visibly determinant $1$.
The permutation $\sigma$ has $s = n/2$ two-cycles if $n$ is even, and $s = (n-1)/2$ two-cycles if $n$ is odd. So indeed
$$
(-1)^{s} = (-1)^{n (n-1)/2}
=
\begin{cases}
(-1)^{n/2} & \text{if $n$ is even, as $(-1)^{n-1} = -1$,}\\
(-1)^{(n-1)/2} & \text{if $n$ is odd, as $(-1)^{n} = -1$.}\\
\end{cases}
$$
