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?

\begin{align}
& \left( \sum_{j=0}^\infty \frac{a^j}{j!} \right) \left( \sum_{k=0}^\infty \frac{b^k}{k!} \right) \\[10pt]
= & \Big(\bullet\Big) \left( \sum_{k=0}^\infty \frac{b^k}{k!} \right) = \sum_{k=0}^\infty \left( \bullet \cdot \frac{b^k}{k!} \right) = \sum_{k=0}^\infty \left(\left( \sum_{j=0}^\infty \frac{a^j}{j!} \right) \frac{b^k}{k!} \right) \\[8pt]
= {} & \sum_{k=0}^\infty \sum_{j=0}^\infty \left( \frac{a^j}{j!} \cdot \frac{b^k}{k!} \right).
\end{align}
This sum includes all the cases where $j+k=0,$ and all cases where $j+k=1$, and all cases where $j+k=2,$ and so on. What are the cases where $j+k=4,$ for example? They are these:
$$
\frac{a^0b^4}{0!4!} + \frac{a^1 b^3}{1!3!} + \frac{a^2b^2}{2!2!} + \frac{a^3 b^1}{3!1!} + \frac{a^4b^0}{4!0!}
$$
That is the same as
\begin{align}
& \frac 1 {4!} \left( \frac{4!}{0!4!} a^0b^4 + \frac{4!}{1!3!} a^1 b^3 + \frac{4!}{2!2!} a^2b^2 + \frac{4!}{3!1!} a^3 b^1 + \frac{4!}{4!0!} a^4b^0 \right) \\[8pt]
= {} & \frac 1 {4!} (a+b)^4.
\end{align}
And similarly for other numbers than $4.$ Thus the sum is
$$
\sum_{\ell=0}^\infty \frac 1 {\ell!} (a+b)^\ell = e^{a+b}.
$$
