Intuitively, we know that the first person has only 1 way to sit since all seats and rotations are identical options at this point. The next person now has a contraint and there are n-1 ways left for them to sit. The next person will have n-2 options and so on following the familiar factorial pattern.
Mathematically, we can write the above as the following noting that first seat is essentially asking how many ways there are for a person to choose 1 seat in n \({n\choose 1}\) which is 1.
\[ {n\choose 1}\times (n-1)\times(n-2)\times(n-3)...3\times 2 \times 1 \\ = 1 \times (n-1)\times(n-2)\times(n-3)...3\times 2 \times 1 \\ = 1 \times (n-1)! = (n-1)! \]