The Huge Matrix

$\begin{aligned} \begin{vmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 2 & 2 & 2 & 2 & 2 \\ 1 & 2 & 3 & 3 & 3 & 3 \\ 1 & 2 & 3 & 4 & 4 & 4 \\ 1 & 2 & 3 & 4 & 5 & 5 \\ 1 & 2 & 3 & 4 & 5 & 12 \end{vmatrix}\end{aligned}$ $=\begin{aligned} \begin{vmatrix} 1-\frac{1}{2} & 0 & 0 & 0 & 0 & 0 \\ 1-\frac{2}{3} & 2-\frac{4}{3} & 0 & 0 & 0 & 0 \\ 1-\frac{3}{4} & 2-\frac{6}{4} & 3-\frac{9}{4} & 0 & 0 & 0 \\ 1-\frac{4}{5} & 2-\frac{8}{5} & 3-\frac{12}{5} & 4-\frac{16}{5} & 0 & 0 \\ 1 & 2 & 3 & 4 & 5 & 5 \\ 1 & 2 & 3 & 4 & 5 & 12 \end{vmatrix}\end{aligned}$ $= \left( 1-\frac{1}{2} \right) \times \left( 2-\frac{4}{3} \right) \times \left( 3-\frac{9}{4} \right) \times \left( 4-\frac{16}{5} \right) \times \begin{aligned} \begin{vmatrix} 5 & 5 \\ 5 & 12 \end{vmatrix}\end{aligned}$ $= \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times 35 = 7$

Note that starting with the second column you can subtract each column by its predecessor without changing the determinant. Doing this we will get triangular matrix with its main diagonal coefficients given by 1 1 1 1 1 7. As the determinant of any triangular matrix can be given by the product of its main diagonal coefficients we get that the given determinant is1x1x1x1x1x7=7