(i) ( n^3 - n )

[ n^3 - n = n(n^2 - 1) = n(n-1)(n+1) = (n+1)n(n-1) = \dfrac{(n+1)!}{(n-2)!}. ]

(Alternatively ( n(n-1)(n+1) ).)

Answer: ( \dfrac{(n+1)!}{(n-2)!} )

(ii) ( n(n-1)(n-2)\cdots(n-r+1) )

[ n(n-1)\cdots(n-r+1) = \dfrac{n!}{(n-r)!}. ]

Answer: ( \dfrac{n!}{(n-r)!} )