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?

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
$$
(-1)^{s} = (-1)^{n (n-1)/2}
=
\begin{cases}
(-1)^{n/2} & \text{if $n$ is even, as $(-1)^{n-1} = -1$,}\\
(-1)^{(n-1)/2} & \text{if $n$ is odd, as $(-1)^{n} = -1$.}\\
\end{cases}
$$
