I'm trying to find the generating function of $\frac{2^k}{k!}$, as there is a $k!$ in the denominator, it must be related to $e^x$, then perhaps, just a simple substitution is needed.
The sequence given by $a_k=\frac{2^k}{k!}$ is $\left\{1,2,2,\frac{4}{3},\frac{2}{3},\frac{4}{15},\frac{4}{45},\frac{8}{315},\frac{2}{315},\frac{4}{2835},\frac{4}{14175}\right\}$, the generating functions I proposed are:
$$e^{2x}=\ldots + \frac{4 x^6}{45}+\frac{4 x^5}{15}+\frac{2 x^4}{3}+\frac{4 x^3}{3}+2 x^2+2 x+1$$
$$e^{2^{x}}=\ldots + 2^x+2^{2 x-1}+\frac{1}{3} 2^{3 x-1}+\frac{1}{3} 2^{4 x-3}+\frac{1}{15} 2^{5 x-3}+\frac{1}{45} 2^{6 x-4}+1$$
I know that the coefficients of $e^{2x}$ are exactly the ones in the sequence of $a_k$ and this would show the answer. The problem is that I spent some time thinking about it, I looked at the answers in the back of the book and the generating function was actually $e^{2x}$. I have the answer, but I don't fully understand what's happening.
EDIT: My problem is that using $2x$ is just like a lucky guess. It could be anything, is there any systematic method that shows precisely that $y$ must be $2x$?
I may be pointing out the obvious, but...
Set $y=2x$. Then $$ 1+y+\frac{y^2}{2!}+\frac{y^3}{3!}+\dots = 1+2x+\frac{(2x)^2}{2!}+\frac{2x^3}{3!}+\dots = 1+2x+\frac{2^2}{2!}x^2+\frac{2^3}{3!}x^3+\dots $$
Set $z=2^x$. Then, $$ 1+z+\frac{z^2}{2}+\frac{z^3}{3!}+\dots = 1+2^x+\frac{2^{2x}}{2!}+\frac{2^{3x}}{3!}+\dots $$ It is possible to use $2^x=e^{x\log 2}$ to transform this further into an ordinary power series, but I think that from these substitutions you can already see why things aren't turning out as you expected.