Find the sum of the infinite series: $ 1 + \frac{1+2}{2!} + \frac{1+2+2^2}{3!} +\frac{1+2+2^2+2^3}{4!}+...$ 

Find the sum of the infinite series
$$ 1 + \frac{1+2}{2!} + \frac{1+2+2^2}{3!} +\frac{1+2+2^2+2^3}{4!}+... ....$$


What I have done let 
$$ S = \underbrace{\frac{1}{1!}}_{\text{1st Term}} + \underbrace{\frac{1+2}{2!}}_{\text{2nd Term}}  +  \underbrace{\frac{1+2+2^2}{3!}}_{\text{3rd Term}}  + \underbrace{\frac{1+2+2^2+2^3}{4!}}_{\text{4th Term}}   +... ....$$
I can see the denominator can be written as such but I'm not sure how to manipulate the numerator?
$$ S = \sum^{\infty}_{n=1} \frac{\text{?}}{n!} $$

The numerator is simply geometric progression,
$$1 + 2 + 2^2 + \cdots + 2^{n-1} = 2^n - 1$$
Therefore,
$$\sum_{n=1}^{\infty} \frac{2^n - 1}{n!} = \sum_{n = 1}^{\infty}\frac{2^n}{n!} - \sum_{n = 1}^{\infty}\frac{1}{n!} = (e^2 - 1) - (e - 1) = e^2 -e$$
