I am trying to figure out how to solve inhomogenous 1. Order Difference Equations and i have been given the following exercise.
$$ y(n+1)=(n+1)y(n)+2^n(n+1)!, \quad y(0)=c, \quad n>0 $$
I use the following formula to find $y(n)$. $$ y(k+1) = \Bigg[\prod_{i=n_0}^{k}a(i)\Bigg]y_0+\sum_{r=n_0}^{k}\Bigg[\prod_{i=r+1}^{k}a(i)\Bigg]g(r) $$ Thus giving me: $$ (1)\qquad y(n) = \Bigg[\prod_{i=0}^{n-1}(i+1)\Bigg]\cdot c + \sum_{r=0}^{n-1}\Bigg[\prod_{i=r+1}^{n-1}(i+1)\Bigg]2^r(r+1)! $$
I know the solution is,
$$ y(n)=n!(2^n+c-1) $$
I'm having problems simplifying (1), you can follow my train of thought in the following deriviation:
$$ (1) \qquad y(n) = \Bigg[\prod_{i=0}^{n-1}(i+1)\Bigg]\cdot c + \sum_{r=0}^{n-1}\Bigg[\prod_{i=r+1}^{n-1}(i+1)\Bigg]2^r(r+1)! $$
$$ (1.a) \qquad y(n) = cn! + \sum_{r=0}^{n-1}\Bigg[\prod_{i=r+1}^{n-1}(i+1)\Bigg]2^r(r+1)! $$
And now i get: $$ (1.b) \qquad y(n) = cn! + \sum_{r=0}^{n-1}n!2^r(r+1)! $$
Where i should get: $$ (1.b*) \qquad y(n) = cn! + \sum_{r=0}^{n-1}n!2^r $$
I simply can't wrap my head around how: $$ \sum_{r=0}^{n-1}\Bigg[\prod_{i=r+1}^{n-1}(i+1)\Bigg]2^r(r+1)!=\sum_{r=0}^{n-1}n!2^r $$