Question 594307
n!-2(n-2)! / (n-2)(n-2)!

n!-2(n-2)!/ (n-2)(n-2)! = (n(n-1)(n-2)! -2(n-2)! ) /  ( (n-2)(n-2)!)
                        = (n(n-1) -2 )/ (n-2)
                        =  (n^2 -n -2)/ (n-2)


  At this point, we have to look for the factors of n^2-n-2, if it exists

Using the quadratic formula, we get
  n^2-n-2=0   ,   --> n= (1- sqrt( 1-4(-2)) )/2 or n = (1+sqrt( 1-4(-2))) /2
                     so n= -1 or n=2

this means the factorization of n^2-n-2 is (n-2)(n+1)
                       

  so,  n!-2(n-2)!/(n-2)(n-2)! = (n+1)(n-2)/ (n-2)
                              = n+1

so the complete simplification of n!-2(n-2)!/(n-2)(n-2)! is  n+1