d(n) = (n − 1) (d(n − 1) + d(n − 2))

a) By considering d(n)−nd(n−1), use the formula above repeatedly to show that d(n)

satisfies the (first order) recurrence relation

d(n) = nd(n - 1) + (-1)^n, for n>=2

b) Use the previous part to show by induction that, for n >= 1, d(n) satisfies the explicit

equation

d(n) = n!(1-(1/1!)+(1/2!)-(1/3!)+(1/4!)-...+((-1)^n)/n!))

I was just about able to prove the formula at the top in a question before this (after a lot of help) which was very difficult, hopefully these 2 questions won't be as hard but i have no idea how to begin. Any help would be great