n! − 2(n − 1)! = (n − 1)(n − 2)(n − 2)!
I don't understand how to even start, an algebraic proof isn't an acceptable answer
The LHS is the number of permutations of such that 1 ends up neither in the first nor in the second position.
( permutations in all, permuations with 1 in the first position, and permutations with 1 in the second position , hence permutations ...)
So let's count this in a different way, first, let's choose a number for the first position, we have choices since we can't choose 1, for the second position we have choices -since we can't choose 1 and we can't the number that we have chosen for the first position.
To complete the rest we have ways, thus: ways in all.