Prove by Structural induction, circular permutations

Question

Prove by Structural Induction:
For a circular permutation of $n$ elements, the number of permutations is $(n-1)!$

How is this done?

Answer 1

Seat one person at a (circular) table: $P(1) = 1 (=0!)$.

Seat the second person at the table. There's only one place for them to go, so $P(2) = 1(=1!)$.

Seat the third person at the table. That person can be seated with Person 1 on his left, or Person 2 on his left. So $P(3) = 2 \cdot P(2) = 2(=2!).$

Seat the fourth person. That person can be seated with any of the three people at the table on his left. So $P(4) = 3 \cdot P(2) = 3 \cdot 2 = 6 = 3!.$

Seat the $n$th person. That person can be seated with any of the already-seated $n-1$ people on his left. So $P(n) = (n-1) \cdot P(n-1) = (n-1)(n-2)\cdots(2)(1) = (n-1)!.$