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?
 A: 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.
A: The expression $x+X$ is meaningless as $x$ and $X$ are quite different things (one is scalar, the other one a matrix), hence it is not
$$
e^{x+X}=1+{x+X}+{(x+X)^2\over 2!}+\cdots 
$$
But the following identity turns out to be true:$$e^{xI+X}=1+{xI+X}+{(xI+X)^2\over 2!}+\cdots$$where $I$ is the identity matrix of the same order of $X$. This is indeed true because of the following implication:$$AB=BA\implies e^{A}e^{B}=e^{A+B}$$
A: \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
$$
\sum_{\ell=0}^\infty \frac 1 {\ell!} (a+b)^\ell = e^{a+b}.
$$
A: 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.
A: You can try the following without resorting to series expansion of $e^x$:
We'll use the following famous limit :
$ e^x= \lim_{n \to \infty} (1+n^{-1})^{nx} $ 
Hence, $e^x\cdot e^X=\lim_{n \to \infty} (1+n^{-1})^{nx}\lim_{n \to \infty} (1+n^{-1})^{nX}=\lim_{n\to \infty} (1+n^{-1})^{n(x+X)}=e^{x+X} $
