\begin{vmatrix} 1 & n & n & \dots & n \\ n & 2 & n & \dots & n \\ n & n & 3 & \dots & n \\ \vdots & \vdots& \vdots & \ddots & \vdots\\ n & n & n & & n\\ \end{vmatrix}
I've the feeling that I should to eliminate with the last row the others ove and after that should I multiply with the elements of the main diagonal.
By Gaussian elimination, $$\begin{vmatrix} 1 & n & n & \dots & n \\ n & 2 & n & \dots & n \\ n & n & 3 & \dots & n \\ \vdots & \vdots& \vdots & \ddots & \vdots\\ n & n & n & \dots & n\\ \end{vmatrix}=\begin{vmatrix} 1 & n & n & \dots & n \\ n-1 & 2-n & 0 & \dots & 0 \\ n-1 & 0 & 3-n & \dots & 0 \\ \vdots & \vdots& \vdots & \ddots & \vdots\\ n-1 & 0 & 0 & \dots & 0\\ \end{vmatrix}$$ so if we expand along the last column, then along the last row, we get that the determinant equals $\color{red}{n!(-1)^{n+1}}$.