<s>[INST] How does $e^x\cdot e^X$ equal $e^{x+X}$? I know that they equal each other, but when I'm trying to prove it, something doesn't match. Please mind the difference between the two equations, one is a lowercase $x$ and the other is an uppercase $x.$ I know that the formula to get $e^x$ is $\frac{x^n}{n!}$. So I apply on $e^x$and it becomes $1+x+\frac{1}{2}x^2+\frac{1}{6}x^3+\cdots$ and the same for $e^X$. This image shows what I'm thinking, and what happens when I multiply them on each other. But I want to go more deeper and show the full equation of the numbers shown, for example, showing $\frac{1}{6}x^3$ and others as well. If I did, there would be a total of 16 numbers shown that are arranged in order of degrees. So I try to prove it like  this picture. But the problem shows when I try to prove it more. I tried proving it more, but $\frac{1}{6}(x+X)^3$ won't work. Than what should I do to make it work? Or what formula should I use?
 [/INST] Lets look at a couple functions. $Y=e^X$ and $y=e^x$.
Then $\ln Y=X$ and $\ln y = x$. $$X+x=\ln (Y)+\ln (y) = \ln (Yy) \Longrightarrow e^{X+x}=Yy=e^Xe^x$$
Ok, so there is a little informal proof using some properties of the natural log.
$$e^z = \sum_{n=0}^{\infty}  \frac{z^n}{n!}$$ (Note: I am not using z as a complex variable, only a stand in for x because it is used so much in the question.)
So, $$e^x=1+x+\frac{x^2}{2} + \frac{x^3}{6}+\frac{x^4}{24}+...+\frac{x^n}{n!}+...$$
$$e^X=1+X+\frac{X^2}{2}+\frac{X^3}{6}+\frac{x^4}{24}+...+\frac{X^n}{n!}+...$$
$$\begin{align*} e^xe^X &= \bigg(1+x+\frac{x^2}{2} + \frac{x^3}{6}+\frac{x^4}{24}+...+\frac{x^n}{n!}+... \bigg)\bigg(1+X+\frac{X^2}{2}+\frac{X^3}{6}+\frac{x^4}{24}+...+\frac{X^n}{n!}...\bigg) \\&= 1\bigg(1+X+\frac{X^2}{2}+\frac{X^3}{6}+\frac{x^4}{24}+...\bigg) + x\bigg(1+X+\frac{X^2}{2}+\frac{X^3}{6}+\frac{x^4}{24}+...\bigg)+ \frac{x^2}{2}\bigg(1+X+\frac{X^2}{2}+\frac{X^3}{6}+\frac{x^4}{24}+...+\frac{X^n}{n!}...\bigg)...\\&= \bigg(1+X+\frac{X^2}{2} + \frac{X^3}{6}...\bigg) + \bigg(x+xX+x\frac{X^2}{2}+x\frac{X^3}{6}...\bigg)+\bigg(\frac{x^2}{2} + \frac{x^2}{2}X + \frac{x^2}{2}\frac{X^2}{2}+\frac{x^2}{2}\frac{X^3}{6}...\bigg)+...\\&= 1+(x+X)+\bigg(\frac{x^2}{2}+xX+\frac{X^2}{2}\bigg)+\bigg(\frac{x^3}{6} + \frac{x^2X}{2}+\frac{X^2x}{2}+\frac{X^3}{6}\bigg)+...\\ &= 1+(x+X)^1 +\frac{(x+X)^2}{2} + \frac{(x+X)^3}{6} + ...\\&= e^{x+X}\end{align*}$$
Another way: $$\begin{align*} e^{x+X}=\sum_{j=0}^{\infty} \frac{1}{j!}(x+X)^j&= \sum_{j=0, 0\le k\le j}^{\infty} \frac{1}{j!} {j\choose k}x^{j-k}X^k \\ &= \frac{1}{0!}\bigg[{0\choose 0}(x^{0-0}X^0)\bigg] + \frac{1}{1!}\bigg[{1\choose 0}(x^{1-0}X^0)+{1\choose 1}(x^{1-1}X^1)\bigg] +\frac{1}{2!}\bigg[{2\choose 0}(x^{2-0}X^0)+{2\choose 1}(x^{2-1}X^1) + {2\choose 2}(x^{2-2}X^2)\bigg] +...\\&= 1+x+X + \frac{x^2}{2} +xX + \frac{X^2}{2} +\frac{x^3}{6} + \frac{x^2X}{2}+\frac{X^2x}{2}+\frac{X^3}{6}+...\\&= \bigg(1+X + \frac{X^2}{2} + \frac{X^3}{6} + ...\bigg)+\bigg(x +xX + x\frac{X^2}{2}+...\bigg) + \bigg(\frac{x^2}{2}+\frac{x^2 X}{2} +\frac{x^2X^2}{2\cdot 2}+...\bigg)+...\\&= \bigg(1+X + \frac{X^2}{2} + \frac{X^3}{6} + ...\bigg)+x\bigg(1+X+\frac{X^2}{2}+...\bigg)+\frac{x^2}{2}\bigg(1+X+\frac{X^2}{2}+...\bigg)+...\end{align*}$$
You can finish it from there.
The exact reason this works is hidden in the way the Taylor series defines a function. I will direct you to this answer, which in my opinion, is a really nice and different way of understanding the intuition behind the Taylor series. The top answer on that page directs to another link of the proof of the Taylor series as well.
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