Problem 706

8. 128 .

Suppose for a certain permutation that satisfies the conditions, the first element on the left is $k(1 \leqslant k \leqslant 8)$. Then, among the remaining seven numbers, the $8-k$ numbers greater than $k$, $k+1, k+2, \cdots, 8$, must be arranged in ascending order; and the $k-1$ numbers less than $k$, $1,2, \cdots, k-1$, must be arranged in descending order (not necessarily adjacent).
In fact, for any number $k+n$ greater than $k$, assume $k+n<8$.
If $k+n+1$ is placed to the left of $k+n$, then the other number $k+n+2$ that differs by 1 from $k+n+1$ must be placed to the left of $k+n+1$; similarly, the other number $k+n+3$ that differs by 1 from $k+n+2$ must be placed to the left of $k+n+2$; $\cdots$. Then the second number in the permutation cannot differ by 1 from $k$, which is a contradiction.
Therefore, $k+n+1$ must be placed to the right of $k+n$.
Similarly, the $k-1$ numbers less than $k$, $1,2, \cdots, k-1$, must be arranged in descending order.

Since when the first element on the left $k$ is determined, there are still seven positions on the right, and any $8-k$ of these positions can be chosen to fill in the numbers greater than $k$ (the remaining $k-1$ positions will fill in the numbers less than $k$), the number of ways to choose is $\mathrm{C}_{7}^{8-k}$; and once the positions are chosen, the method of filling in the numbers is uniquely determined.
Therefore, the total number of permutations is
$$\sum_{k=1}^{8} \mathrm{C}_{7}^{8-k}=\sum_{k=0}^{7} \mathrm{C}_{7}^{j}=2^{7} .$$