Evalute big determinant Today in exam I tried to evaluate this determinant but failed, only somehow "guessed" the answer I got here. Now in home I've managed to find something intuitive, just want to know whether the approach is correct, and is there more faster way exist. Given determinant
$$\det\begin{vmatrix}
1 & 2 & 3 & ... & n-2 & n-1 & n\\ 
2 & 3 & 4 & ... & n-1 & n & n\\ 
3 & 4 & 5 & ... & n & n & n \\ 
. & . & . & . & . & . &. \\
n & n & n & ... & n & n & n
\end{vmatrix}$$

First thing I did, was rearranging rows. I remember from another problem, where I used to evaluate determinant of matrix of this kind $\det\begin{vmatrix}
0 & 0 ... & 0 & 1\\ 
0 & 0 ... & 1 & 0\\ 
0 & 0 ... & 0 & 0\\ 
. & . & . & . \\
1 & 0 ... & 0 & 0
\end{vmatrix}$ is $(-1)^{\frac{n(n-1)}{2}}*\det\begin{vmatrix}
1 & 0 ... & 0 & 0\\ 
0 & 1 ... & 0 & 0\\ 
0 & 0 ... & 0 & 0\\ 
. & . & . & . \\
0 & 0 ... & 0 & 1
\end{vmatrix}=(-1)^{\frac{n(n-1)}{2}}$.
So it becomes
$$(-1)^{\frac{n(n-1)}{2}}\det\begin{vmatrix}
n & n & n & ... & n & n & n \\
. & . & . & . & . & . &. \\
3 & 4 & 5 & ... & n & n & n \\ 
2 & 3 & 4 & ... & n-1 & n & n\\ 
1 & 2 & 3 & ... & n-2 & n-1 & n\\ 
\end{vmatrix}$$
And then I transposed it
$$(-1)^{\frac{n(n-1)}{2}}\det\begin{vmatrix}
n & n-1 & n-2 & ... & 3 & 2 & 1 \\
. & . & . & . & . & . &. \\
n & n & n & ... & n & n-1 & n-2 \\ 
n & n & n & ... & n & n & n-1\\ 
n & n & n & ... & n & n & n\\ 
\end{vmatrix}$$
and tried to subtract first row from all.
$$(-1)^{\frac{n(n-1)}{2}}\det\begin{vmatrix}
n & n-1 & n-2 & ... & 3 & 2 & 1 \\
0 & 1 & 1 & ... & 1 & 1 & 1 \\ 
0 & 1 & 2 & ... & 2 & 2 & 2\\ 
. & . & . & . & . & . &. \\
0 & 1 & 2 & ... & n-3 & n-2 & n-1\\ 
\end{vmatrix}$$
next step is subtracting second row from others below.
$$(-1)^{\frac{n(n-1)}{2}}\det\begin{vmatrix}
n & n-1 & n-2 & ... & 3 & 2 & 1 \\
0 & 1 & 1 & ... & 1 & 1 & 1 \\ 
0 & 0 & 1 & ... & 1 & 1 & 1\\ 
. & . & . & . & . & . &. \\
0 & 0 & 1 & ... & n-4 & n-3 & n-2\\ 
\end{vmatrix}$$
doing this for finite n we'll get
$$(-1)^{\frac{n(n-1)}{2}}\det\begin{vmatrix}
n & n-1 & n-2 & ... & 3 & 2 & 1 \\
0 & 1 & 1 & ... & 1 & 1 & 1 \\ 
0 & 0 & 1 & ... & 1 & 1 & 1\\ 
. & . & . & . & . & . &. \\
0 & 0 & 0 & ... & 0 & 0 & 1\\ 
\end{vmatrix}$$
So my final solution is $n*(-1)^{\frac{n(n-1)}{2}}$. Do you see any mistakes? And maybe there's more easier approach? thanks.
 A: First we can subtract from each row the following row
$$D=
\begin{vmatrix}
1 & 2 & 3 & \ldots & n-2 & n-1 & n\\
2 & 3 & 4 & \ldots & n-1 & n & n\\
3 & 4 & 5 & \ldots & n & n & n \\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
n & n & n & \ldots & n & n & n
\end{vmatrix}
=
\begin{vmatrix}
-1&-1 &-1 & \ldots &-1 &-1 & 0\\
-1&-1 &-1 & \ldots &-1 & 0 & 0\\
-1&-1 &-1 & \ldots & 0 & 0 & 0\\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
n & n & n & \ldots & n & n & n
\end{vmatrix}$$
We can "pull out" $n$ from the last row, and then add the last row to all rows above it (to cancel out all $-1$'s)
$$D=n
\begin{vmatrix}
-1&-1 &-1 & \ldots &-1 &-1 & 0\\
-1&-1 &-1 & \ldots &-1 & 0 & 0\\
-1&-1 &-1 & \ldots & 0 & 0 & 0\\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
1 & 1 & 1 & \ldots & 1 & 1 & 1
\end{vmatrix}=
n
\begin{vmatrix}
 0& 0 & 0 & \ldots & 0 & 0 & 1\\
 0& 0 & 0 & \ldots & 0 & 1 & 1\\
 0& 0 & 0 & \ldots & 1 & 1 & 1\\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
1 & 1 & 1 & \ldots & 1 & 1 & 1
\end{vmatrix}
$$
Now we only need to subtract each row from the row following it to get
$$
D=
n
\begin{vmatrix}
 0& 0 & 0 & \ldots & 0 & 0 & 1\\
 0& 0 & 0 & \ldots & 0 & 1 & 1\\
 0& 0 & 0 & \ldots & 1 & 1 & 1\\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
1 & 1 & 1 & \ldots & 1 & 1 & 1
\end{vmatrix}=
n
\begin{vmatrix}
 0& 0 & 0 & \ldots & 0 & 0 & 1\\
 0& 0 & 0 & \ldots & 0 & 1 & 0\\
 0& 0 & 0 & \ldots & 1 & 0 & 0\\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
1 & 0 & 0 & \ldots & 0 & 0 & 0
\end{vmatrix}=n(-1)^{\frac{n(n-1)}2}$$
(We got the matrix for which you already know the determinant. It can be obtained, for example, using several swaps of neighbouring rows.) 
