I found this in the prologue of Rudin's book. The formula tries to relate this exponential addition using power series. It is clear that if we start from the right we can write it as a sum of binomial expansion terms divided by factorial.
If we come from the left, I am a bit vexed. Working out the terms by hand
(1 + a + a^2/2 + a^3/6 + a^4/24 + a^5/120 + ...)(1 + b + b^2/2 + b^3/6 + b^4/24 + b^5/120 + ...) =
1 + b + b^2/2 + b^3/6 + b^4/24 + b^5/120 + ... +
a + ab + ab^2/2 + ab^3/6 + ab^4/24 + ab^5/120 + ... +
(a^2 + a^2b + a^2b^2/2 + a^2b^3/6 + a^2b^4/24 + a^2b^5/120 + ... +)/2
(a^3 + a^3b + a^2b^3/2 + a^3b^3/6 + a^3b^4/24 + a^3b^5/120 + ... +)/6
Here I am just factoring each term of exp(a) series with the entire series of exp(b).
(a+b)^0/1 = 1
(a+b)^1/1 = a + b
(a+b)^2/2 = a^2/2 + ab + b^2/2
(a+b)^3/6 = a^3/6 + a^2*b/2 + a*b^2/2 + b^3/6
I am able to visually see that if we take diagonals each one works out to a binomial expansion divided by factorial.
What is the actual mathematical concept for this? It looks possibly related to Pascal's triangle.
Best Answer
Let us start writing $\exp(x+y)$ as power series, and use the binomial theorem to expand $(x+y)^n$: $$\begin{align} e^{x+y} &= \sum_{n\geqslant 0}\frac1{n!}(x+y)^n\\ &= \sum_{n\geqslant 0}\frac1{n!}\sum_{j=0}^n \binom nj x^j y^{n-j}\\ \end{align}$$ Now denote $f(x)[x^k]$ as the coefficient at $x^k$ in the power series expansion of $f(x)$. Applying $[x^k]$ to the series above, the inner sum reduces to just one term at most due to $j\stackrel!=k$: $$\begin{align} e^{x+y}[x^k] &= \sum_{n\geqslant k}\frac1{n!} \binom nk y^{n-k}\\ &\stackrel{1.}= \sum_{n\geqslant 0}\frac1{(n+k)!} \binom {n+k}k y^n\\ &\stackrel{2.}= \frac 1{k!} \sum_{n\geqslant 0}\frac{k!}{(n+k)!} \frac{(n+k)!}{k!n!} y^n\\ &= \frac 1{k!} \underbrace{\sum_{n\geqslant 0}\frac{1}{n!}y^n}_{\displaystyle{=e^y}}\\ \end{align}$$
Replace the summation index as $n\mapsto n+k$.
Extends by $k!$, and expands the binomial coefficient.
The last line is what we wanted to show because $$\begin{align} e^x e^y &= \left(\sum_{k\geqslant 0}\frac{x^k}{k!}\right)\left(\sum_{n\geqslant 0}\frac{y^n}{n!}\right)\\ &= \sum_{k\geqslant 0}\left(\frac{1}{k!}\sum_{n\geqslant 0}\frac{y^n}{n!}\right)x^k\\ \end{align}$$ $$\begin{align} \end{align}$$ $$\begin{align} \end{align}$$