Squaring Infinite Series Expansion Of e^x $Fact$:$$\lim\limits_{n \to \infty}\frac{x^0}{0!}+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots+\frac{x^n}{n!}=e^x$$
so
$$\lim\limits_{n \to \infty}\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots+\frac{1}{n!}=e$$
also
$$\lim\limits_{n \to \infty}e^2=\frac{2^0}{0!}+\frac{2}{1!}+\frac{4}{2!}+\frac{8}{3!}+\dots+\frac{2^n}{n!}$$
can I safely say...
$$L.H.S=(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots)^2=\frac{2^0}{0!}+\frac{2}{2!}+\frac{4}{2!}+\frac{8}{3!}+\dots=R.H.S$$
if yes, how to prove L.H.S=R.H.S (without using the above fact)?
It is not a problem from textbook I was just thinking about it. 

Since $e^ne^m = e^{n+m}$ you can simply let $x = 2t$, giving you 
$$(e^{t})^2 = e^{2t} = 1 + 2t + \frac{4t^2}{2!}+\frac{8t^3}{3!} + \cdots$$
Now, let $t = 1$ and you have 
$$e^2 = 1 + 2 + \frac{4}{2!} + \frac{8}{3!} + \cdots = (1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots )^2 = (e^1)^2.$$
