Here we look for what does not change.(method of invariants). We replace x and y by xy-x-y+2= (x-1)(y-1)+1
So, if we subtract 1, from each number and multiply them , the product will be constant in any step.
Initially the value of the product was (n-1)!
So when we are left with only two numbers call a and b, we must have (a-1)(b-1) = (n-1)!
In the last step we replace a,b by (a-1)(b-1) + 1 = (n-1)! + 1
Hence we will always be left with (n-1)! + 1