First, that original representation you have should not have $(n-2)!$, but just $(n-2)$.
The "..." is just to show you are continuously multiplying by the next lowest integer until you reach $1$. For someone who has never learned about the factorial that second notation could be confusing (it may seem obvious to most to stop at $1$, and not $0$ or negative numbers, but it's good to be clear in the beginning).
For the case $n=3$ as you asked, you wouldn't list $3$ twice in the product, one of the $(n-something)$ would be your $3$, and you would multiply all lesser integers (one time each) until you reach $1$ (which means you would just multiply $3$, $2$ and $1$).
So $3! = 3 \times 2 \times 1$ is all you need and when you see
$n! = n \times (n-1) \times (n-2) \times ... \times 3 \times 2 \times 1$
it just means take your $n$ and multiply it with all the smaller integers less than $n$. The "..." is just a placeholder for all the other integers less than $n$. For $n=3$ your $n$, $(n-1)$, and $(n-2)$ are already your $3$, $2$, and $1$ values, so you don't list them twice.