7. As shown in Figure 1, under the rules of Chinese chess, the "pawn" at point $A$ can reach point $B$ through a certain path (before crossing the river, the pawn can only move to the adjacent intersection directly in front of it each step; after crossing the river, it can move to the adjacent intersections in front, to the left, or to the right each step, but it cannot move backward. The "river" refers to the area between the 5th and 6th horizontal lines on the chessboard in Figure 1). During the pawn's movement, if each intersection on the chessboard is not visited more than once by the pawn, such a path is called a "non-repeating path." Therefore, the number of different non-repeating paths is $\qquad$
7. 6561.

Assume the chessboard has 10 horizontal lines from bottom to top, sequentially labeled as rows 1, 2, ..., 10, and 9 vertical lines from left to right, sequentially labeled as columns 1, 2, ..., 9. For example, point A is located at the 4th row and 5th column.

Notice that, during the movement of the pawn from point A to B, it cannot move downward. Therefore, the step from row $i$ to row $i+1$ is unique. If the starting and ending points of this step are in column $j$, we denote $a_{i}=j (4 \leqslant i \leqslant 9, 1 \leqslant j \leqslant 9)$.
According to the rules, it is easy to see that
$$
a_{4}=a_{5}=5, a_{6}, a_{7}, a_{8}, a_{9} \in\{1,2, \cdots, 9\} .
$$

There are $1^{2} \times 9^{4}=6561$ such ordered arrays $\left(a_{4}, a_{5}, \cdots, a_{9}\right)$, and each array corresponds one-to-one to a unique non-repeating path from point A to B. This is because, after the pawn reaches row $i (i=4,5, \cdots, 9)$, there is exactly one way to move to column $a_{i}$ in row $i$ (otherwise, the path would repeat), and then move forward to row $i+1$. Therefore, the total number of different non-repeating paths is 6561.