Since the graph is complete, any permutation starting with a fixed vertex gives an (almost) unique cycle (the last vertex in the permutation will have an edge back to the first, fixed vertex), except for the fact that if we visit the vertices of the cycle in reverse order, it is really the same cycle. Thus the number of cycles is half of the number of permutations of $(n-1).$
There are $(n-1)!$ permutations of the non-fixed vertices, and half of those are the reverse of another, so there are $\dfrac{(n-1)!}{2}$ distinct Hamiltonian cycles in the complete graph of $n$ vertices.
$$\left(\textbf{OR}\right)$$
The number of ways to arrange $n$ distinct objects along a fixed circle is $(n-1)!.$ If clock-wise and anti-clockwise cycle is same then we divide total permutations with $2$. For example two cycles $123$ and $321$ both are same because they are reverse of each other.
For all $n\geq 3$, the number of distinct Hamilton Cycles in a complete graph $K_{n}$ is $\dfrac{(n-1)!}{2}.$
The number of Hamiltonian cycles in $K_{8} = \dfrac{(8-1)!}{2}=\dfrac{7!}{2} = 2520$.
So, the correct answer is $2520$.
