Sum of Sum of Squares?

Since $1+2+2^2+\cdots+2^n=2^{n+1}-1,$ the given expression changes to $\sum_{k=1}^\infty \dfrac{2^k-1}{k!}=\left(\sum_{k=1}^\infty \dfrac{2^k}{k!}\right)-\left(\sum_{k=1}^\infty \dfrac{1}{k!}\right)=\boxed{e^2-e}$ which is the desired value.

An alternative way to do this would be the Taylorseries $e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$ .

The above expression is equal to $1+\frac{3}{2!}+\frac{7}{3!}+\frac{15}{4!}+... = 1+\frac{2^2-1}{2!}+\frac{2^3-1}{3!}+\frac{2^4-1}{4!}+...$

Now we split it up like this: $1+2+\frac{2^2}{2!}+\frac{2^3}{3!}+\frac{2^4}{4!}+...-(1+1+\frac{1^2}{2!}+\frac{1^3}{3!}+\frac{1^4}{4!}+...)=\boxed{e^2-e}$

$1 + \dfrac{1+2}{2!} + \dfrac{1+2+2^2}{3!} + \dfrac{1+2+2^2+2^3}{4!} + \cdots = \\ 1 + \dfrac{2^2-1}{2!} + \dfrac{2^3-1}{3!} + \dfrac{2^4-1}{4!} + \cdots = \\ 1 - \left( \dfrac{1}{2!} + \dfrac{1}{3!} + \dfrac{1}{4!} + \cdots \right) + \left( \dfrac{2^2}{2!} + \dfrac{2^3}{3!} + \dfrac{2^4}{4!} + \cdots \right) = \\ 1 - (e - 2) + (e^2 - 3) =\\ e^2 - e$