Calculate the determinant
$$ \begin{vmatrix} 1 & 2 & 3 & \cdots & n-1 & n \\ 2 & 3 & 4 & \cdots & n & 1 \\ 3 & 4 & 5 & \cdots & 1 & 2 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ n-1 & n & 1 & \cdots & n-3 & n-2 \\ n & 1 & 2 & \cdots & n-2 & n-1 \end{vmatrix} $$
I know the answer is $(-1)^\frac{n(n-1)}{2}\frac{n^{n-1}(n+1)}{2}$, and it is possible to solve with a tedious amount of linear elementary transformations, i.e. first you add other columns to the first column, and then subtract the first column from other columns, blah blah. However, is there a cleverer and faster way?
