I came across this answer on a very similar question which says:
Total (non-distinct) Hamiltonian circuits in complete graph $K_n$ is $(n−1)!$
This follows from the fact that starting from any vertex we have $n−1$ edges to choose from first vertex, $n−2$ edges to choose from second vertex, $n−3$ to choose from the third and so on. These being independent choices, we get $(n−1)!$ possible number of choices.
I have seen many textbooks which mention the exact same reasoning, but what I do not understand is the fixed choice of the starting vertex.
Since the starting vertex itself can be chosen in $n$ ways, we should have $n!$ total Hamiltonian circuits and not $(n-1)!$.
Why is this not the case?
I will try to merge my discussion with Matthew Leingang in the comments into an answer.
Consider the complete graph on vertices $\{1,2,3\}$. Here $1 \sim 2 \sim 3 \sim 1$ and $2 \sim 3 \sim 1 \sim 2$ are equivalent circuits.
Now, why does it work?
Say our starting vertex is $1$ and we either select the edge to $2$ or $3$. Without the loss of generality, let's say we select $1 \sim 2$. Now we have a circuit which is $1 \sim 2 \sim 3 \sim 1$.
Now, what would happen if we didn't have starting vertex as $1$? We could've had a circuit $2 \sim 1 \sim 3 \sim 2$. However, this is equivalent to $1 \sim 3 \sim 2 \sim 1$ which would've been there had we considered the first edge to be $1 \sim 3$.
Conclusion
The method of choosing an initial vertex first will count each circuit $n$ times instead of once. So, the total number of distinct circuits is:
$n!$ $/$ $n$ $=$ $(n−1)!$