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?

Perhaps you're confused about the arrangement. We should get $e^xe^x=e^{2x}$, which can be written $$e^{2x} = 1 +(2x) + \tfrac{(2x)^2}{2!} + \tfrac{(2x)^3}{3!}+\cdots$$
$$=1 + 2x + \tfrac{4x^2}{2!} + \tfrac{8x^3}{3!} + \cdots$$
So let's multiply series:
$$e^x\cdot e^x = \left(1 + x + \tfrac{x^2}{2!} + \tfrac{x^3}{3!} + \cdots\right)\cdot\left(1 + x + \tfrac{x^2}{2!} + \tfrac{x^3}{3!} + \cdots\right)$$
$$= 1\cdot\left(1 + x + \tfrac{x^2}{2!} + \tfrac{x^3}{3!} + \cdots\right) + x\cdot\left(1 + x + \tfrac{x^2}{2!} + \tfrac{x^3}{3!} + \cdots\right) + \tfrac{x^2}{2!}{\cdot\left(1 + x + \tfrac{x^2}{2!} + \tfrac{x^3}{3!} + \cdots\right)}+\cdots$$
$$= (1) + (1\cdot x + x\cdot 1) + (1\cdot \tfrac{x^2}{2!} + x\cdot x + \tfrac{x^2}{2!}\cdot 1) + \cdots $$
$$= 1 + (2)x + (\tfrac1{2!} + 1 + \tfrac1{2!})x^2 + \cdots$$
$$=1 + 2x + \tfrac4{2!}x^2 + \cdots$$
You collect the products of a fixed degree $n$ as
$$1\cdot \tfrac{x^n}{n!} + x\cdot\tfrac{x^{n-1}}{(n-1)!} + \tfrac{x^2}{2!}\cdot\tfrac{x^{n-2}}{(n-2)!} + \cdots + \tfrac{x^{n-2}}{(n-2)!}\cdot \tfrac{x^2}{2!} +\tfrac{x^{n-1}}{(n-1)!}\cdot x + \tfrac{x^{n}}{n!}\cdot 1$$
$$= (\tfrac1{n!} + \tfrac{n}{n!} + \tfrac{n(n-1)}{n!} + \tfrac{n(n-1)(n-2)}{n!} + \cdots)x^n$$
Your job is to show that this is $\tfrac{2^n}{n!}x^n$ (expand $(1+1)^n)$ using the binomial formula).

Retry:
I think you are doing it right, you just didn't collect all terms correctly.
You are really just using the distributive law, like
$$(a + b + c+\cdots)(\textrm{terms}) = a\cdot(\textrm{terms}) + b\cdot(\textrm{terms}) + c\cdot(\textrm{terms})+\cdots$$
For each of the products on the RHS, you need to look for results of the same degree (we're really looking at the exponential series here, of course).

*

*Constant terms only occur as $1\cdot 1$, in the first product on the RHS.


*Degree 1 terms occur as $1\cdot x$ or $x\cdot 1$ (in the first and second products on the RHS).


*Degree 2 terms occur as $1\cdot x^2$, $x\cdot x$, or $x^2\cdot 1$ (in the first, second, and third products on the RHS).


*Degree 3 terms occur as $1\cdot x^3$, $x\cdot x^2$, $x^2\cdot x$, or $x^3\cdot 1$  (in the 1st, 2nd, 3rd, and 4th products on the RHS).
And so on.
