$1 \cdot 33$ The integers $1,2, \cdots, n$ are arranged in a permutation such that: each number is either greater than all the numbers before it, or less than all the numbers before it. How many such permutations are there?
[Solution] Let the number of permutations we are looking for be $a_{n}$.
When $n=1$, there is only the number 1, so obviously $a_{1}=1$.
For $n \geqslant 2$, if the number $n$ is placed in the $i$-th position, then the $n-i$ numbers after it are completely determined, i.e., they can only be $n-i, n-i-1, \cdots, 1$, and the $i-1$ numbers before it have $a_{i-1}$ ways to be arranged. Considering the different positions of $n$, we must have
$$
a_{n}=1+a_{1}+a_{2}+\cdots+a_{n-1} .
$$

From $a_{1}=1$, we get
$$
\begin{array}{l}
a_{2}=1+a_{1}=2, \\
a_{3}=1+a_{1}+a_{2}=4=2^{2}, \\
a_{4}=1+a_{1}+a_{2}+a_{3}=8=2^{3} .
\end{array}
$$

From this, we can conjecture
$$
a_{n}=2^{n-1} .
$$

This conclusion is not difficult to prove by mathematical induction.