【Example 4】There are $n$ money boxes, each with one key, and the keys are not interchangeable. If a key is randomly placed in each box, and then all the boxes are locked. Now, you are allowed to break open one box. How many ways are there to place the keys so that once one box is broken open, all the boxes can be subsequently opened with the keys?
In order to be able to successively open all the boxes with keys after breaking open one box, the key to another box $i_{1}$ must be placed in the box that is broken open, and in box $i_{1}$, the key to box $i_{2}$ must be placed, and in box $i_{2}$, the key to box $i_{3}$ must be placed, and so on, covering every box, until the key to the box $i_{n-1}$ is placed in box $i_{n-2}$, and the key to the box that was broken open, or the key to box $i_{n}$, is placed in box $i_{n-1}$. This is equivalent to a circular permutation of $1,2, \cdots, n$, because it is sufficient to arrange $i_{1}, i_{2}, \cdots, i_{n}$ in a circle in a counterclockwise direction. Conversely, for each circular permutation of $1,2, \cdots, n$, a method of placing keys that allows all boxes to be successively opened can also be corresponded. First, let the $n$ boxes be seated in order along the circle, then place the key to the box below (determined in a counterclockwise direction) in each box, thus obtaining a method of placing keys that allows all boxes to be successively opened. Therefore, the number of ways to place keys that allow all boxes to be successively opened is exactly equal to the number of circular permutations of $1,2, \cdots, n$, which is a total of $(n-1)$! ways.