139 In an $8 \times 8$ grid of a chessboard, color the squares using 8 different colors, with each square being one color, such that adjacent squares (sharing an edge) are colored differently, and all eight colors appear in each row. The number of different coloring methods is $\qquad$ .
$1398! \times 14833^{7}$. The first row has 8! coloring methods. After coloring the first row, then color the second row. By the principle of inclusion-exclusion, the number of coloring methods is
$$
\begin{aligned}
& 8! - C_{8}^{1} \times 7! + C_{8}^{2} \times 6! - C_{8}^{3} \times 5! + C_{8}^{4} \times 4! - \\
& C_{8}^{5} \times 3! + C_{8}^{6} \times 2! - C_{8}^{7} \times 1! + C_{8}^{8} \\
= & 8! \left( \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \frac{1}{6!} - \frac{1}{7!} + \frac{1}{8!} \right) \\
= & 14833 \text{ (methods). }
\end{aligned}
$$

Similarly, coloring the third row also has 14833 methods. And so on, there are a total of $8! \times 14833^{7}$ methods.