Question

D(1) = 1,d(n) = n.d(n-1) for n ≥ 2

Answer 1

Assuming you're solving for
`d(n)`, we have by substitution

`d(n) = n d(n-1) \implies d(n-1) = (n-1) d(n-2) \\\\ ~~~~ \implies d(n) = n(n-1)d(n-2)`

`d(n) = n d(n-1) \implies d(n-2) = (n-2) d(n-3) \\\\ ~~~~ \implies d(n) = n(n-1)(n-2)d(n-3)`

`d(n) = n d(n-1) \implies d(n-3) = (n-3) d(n-4) \\\\ ~~~~ \implies d(n) = n(n-1)(n-2)(n-3) d(n-4)`

and so on. After
`k` iterations of this, we have the pattern

`d(n) = n(n-1)(n-2)\cdots(n-(k-3))(n-(k-2))(n-(k-1)) d(n-k)`

so that after
`k=n-1` iterations,

`d(n) = n(n-1)(n-2)\cdots(n-(n-3))(n-(n-2)) d(n-(n-1)) \\\\ ~~~~= n(n-1)(n-2)\cdots3\cdot2\,d(1) \\\\ ~~~~ = \boxed{n!}`