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 $\displaystyle \{1,...,n\}$ such that 1 ends up neither in the first nor in the second position.
( $\displaystyle n!$ permutations in all, $\displaystyle (n-1)!$ permuations with 1 in the first position, and $\displaystyle (n-1)!$ permutations with 1 in the second position , hence $\displaystyle n!-2(n-1)!$ permutations ...)
So let's count this in a different way, first, let's choose a number for the first position, we have $\displaystyle (n-1)$ choices since we can't choose 1, for the second position we have $\displaystyle (n-2)$ 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 $\displaystyle (n-2)!$ ways, thus: $\displaystyle (n-1)\cdot (n-2)\cdot (n-2)!$ ways in all.