I don't know much about math though I want to learn everything about it. I was just in the wikipedia page of the Euler's number which says
$e=\displaystyle\sum_{n=0}^{\infty}\frac{1}{n!}$
I also know something about the Exponential function $f(x)=e^x$
So what I just done is multiplying $\displaystyle\sum_{n=0}^{\infty}\frac{1}{n!}$ by itself to get $e^2$
and I got something like this: $e+\frac{e}{2}+\frac{e}{6}+\frac{e}{24}+\frac{e}{120}...$
and then i figured out that $e^x$ equals $\displaystyle\sum_{n=0}^{\infty}\frac{e^{x-1}}{n!}$
Is this a known thing? Sorry for wasting your time if it is, I just want to know more about these things.
If this all started because you were squaring the series $\sum_n\tfrac{1}{n!}$, you probably want to see how we show the exponential series satisfies $\exp(x+y)=\exp x\exp y$. It follows by the binomial theorem:$$\begin{align}\exp(x)\exp(y)&=\sum_{k,\,l\ge0}\frac{x^ky^l}{k!l!}\\&=\sum_{k,\,l\ge0}\frac{\binom{k+l}{k}x^ky^l}{(k+l)!}\\&\stackrel{\star}{=}\sum_{n\ge0}\frac{1}{n!}\sum_{k=0}^n\binom{n}{k}x^ky^{n-k}\\&=\sum_{n\ge0}\frac{(x+y)^n}{n!}\\&=\exp(x+y).\end{align}$$In particular, $\star$ defines $n:=k+l$ and rearranges a double sum by the Fubini principle, exploiting the fact that $\tfrac{1}{n!}$ decays superexponentially, and therefore there are no convergence issues. (The above principle's equivalent on integrals is what we would normally intend by "Fubini's theorem".)