<s>[INST] 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. 
 [/INST] Let $$f(z) = \sum_{n = 0}^{\infty}\frac{z^{n}}{n!}\tag{1}$$ then it is easy to see that the series on right is absolutely convergent for all values of $z \in \mathbb{C}$ and hence the function $f(z)$ is well defined for all complex values of $z$. Using Cauchy's product rule for multiplication of infinite series we see that $$f(z)f(w) = \sum_{n = 0}^{\infty}\frac{z^{n}}{n!}\cdot\sum_{n = 0}^{\infty}\frac{w^{n}}{n!} = \sum_{n = 0}^{\infty}c_{n}\tag{2}$$ where $$c_{n} = \sum_{k = 0}^{n}\frac{z^{n - k}}{(n - k)!}\cdot\frac{w^{k}}{k!} = \frac{1}{n!}\sum_{k = 0}^{n}\binom{n}{k}z^{n - k}w^{k} = \frac{(z + w)^{n}}{n!}\tag{3}$$ and hence from $(2), (3)$ we get $$f(z)f(w) = \sum_{n = 0}^{\infty}\frac{(z + w)^{n}}{n!} = f(z + w)\tag{4}$$ for all complex $z, w$. Putting $z = w = 1$ we get $$f(1)\cdot f(1) = f(2)$$ which is the result asked in question.
Using $(4)$ and little bit of algebra it is possible to prove that $\{f(z)\}^{n} = f(nz)$ for all complex $z$ and rational $n$. Putting $z = 1$ and replace $n$ by $x$ we see that $f(x) = \{f(1)\}^{x}$ where $x$ is rational or in grand fashion $$\left(1 + \frac{1}{1!} + \frac{1}{2!} + \cdots\right)^{x} = 1 + x + \frac{x^{2}}{2!} + \cdots$$ where $x$ is rational (this is the fact you mention in the beginning of your question).
</s>