Proving that one infinite series is equal to another. This is a question I've been messing with all day and still cannot figure out how to prove it. I have tried the ratio test which I assumes would the be the best idea but I need a little more of an idea of how to approach the problem. Here it is: 
Prove that:
$$\sum_{n=0}^{\infty} \frac{2^{n}x^{n}}{n!} = {e^{2}} \left( \sum_{n=0}^{\infty}\frac{(x-1)^n}{n!}\right)^2$$
 A: $e^x=\sum_{n=0}^{\infty}\dfrac{x^n}{n!}$
So, $\sum_{n=0}^{\infty}\dfrac{2^nx^n}{n!}=e^{2x}$
and,
$e^2\sum_{n=0}^{\infty}\dfrac{(x-1)^n}{n!}=e^2e^{x-1}=e^{x+1}$  So this is not true for most values of $x$. If it was $e^2\Big(\sum_{n=0}^{\infty}\dfrac{(x-1)^{n}}{n!}\Big)^2$ instead, it would work.
A: $$
\begin{align}
\sum_{n=0}^\infty\frac{(x-1)^n}{n!}
&=\sum_{n=0}^\infty\sum_{k=0}^n\binom{n}{k}\frac{x^k(-1)^{n-k}}{n!}\\
&=\sum_{n=0}^\infty\sum_{k=0}^n\frac{x^k(-1)^{n-k}}{k!(n-k)!}\\
&=\sum_{k=0}^\infty\sum_{n=k}^\infty\frac{x^k(-1)^{n-k}}{k!(n-k)!}\\
&=\sum_{k=0}^\infty\sum_{n=0}^\infty\frac{x^k(-1)^n}{k!\,n!}\\
&=\sum_{k=0}^\infty\frac{x^k}{k!}\sum_{n=0}^\infty\frac{(-1)^n}{n!}\\
&=\frac1e\sum_{k=0}^\infty\frac{x^k}{k!}\tag{1}
\end{align}
$$
$$
\begin{align}
\left(\sum_{k=0}^\infty\frac{x^k}{k!}\right)^2
&=\sum_{k=0}^\infty\frac{x^k}{k!}\sum_{n=0}^\infty\frac{x^n}{n!}\\
&=\sum_{k=0}^\infty\sum_{n=0}^\infty\frac{x^k}{k!}\frac{x^n}{n!}\\
&=\sum_{k=0}^\infty\sum_{n=k}^\infty\frac{x^k}{k!}\frac{x^{n-k}}{(n-k)!}\\
&=\sum_{n=0}^\infty\sum_{k=0}^n\frac{x^k}{k!}\frac{x^{n-k}}{(n-k)!}\\
&=\sum_{n=0}^\infty\sum_{k=0}^n\binom{n}{k}\frac{x^n}{n!}\\
&=\sum_{n=0}^\infty\frac{2^nx^n}{n!}\tag{2}
\end{align}
$$
Squaring $(1)$ and applying $(2)$ yields
$$
\sum_{n=0}^\infty\frac{2^nx^n}{n!}=e^2\left(\sum_{n=0}^\infty\frac{(x-1)^n}{n!}\right)^2\tag{3}
$$
