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The goal is to prove that

$\exp(x+y)\exp(x)\exp(y)$,

or in another form

$e^{(x+y)}=e^x+e^y$.

1. Show that when you fully expand $\exp(x)\cdot\exp(y)$, each term has the form $\frac{x^k\cdot y^m}{k!\cdot m!}$.

From Cauchy product,
$$\left(\sum_{i=0}^\infty a_i\right) \cdot \left(\sum_{j=0}^\infty b_j\right) = \sum_{k=0}^\infty c_k$$
where, $c_k=\sum_{l=0}^k a_l b_{k-l}$

\begin{align} \nonumber\exp(x)\cdot\exp(y)&=\sum_{n=k+m}^{\infty}\frac{x^{n}}{n!}\sum_{m}^{\infty}\frac{y^{m}}{m!}\\ &=\sum_{k}^{\infty}\sum_{m}^{k}\frac{x^{k}}{k!}\frac{y^{m}}{m!} \end{align}

1. Show that when you expand $\exp(x+y)$, each term has the form $\frac{1}{n!}{n\choose k}x^ky^{n-k}$.

From Binomial theorem,
$$(x+y)^{n}=\sum _{k=0}^{n}{n \choose k}x^{n-k}y^{k}=\sum _{k=0}^{n}{n \choose k}x^{k}y^{n-k}$$

\begin{align} \nonumber\exp(x+y)&=\sum_n\frac{(x+y)^n}{n!}\\ \nonumber&=\sum_{n}\frac{1}{n!}\sum_{n=k+m} {n \choose k} x^ky^m \\ &= \sum_{n=k+m} \frac{1}{n!}{n\choose k}x^ky^{n-k} \end{align}

1. Compare the two results above to explain why $\exp(x+y)=\exp(x)\cdot\exp(y)$.

Compare (1) and (2), as ${n\choose k}=\frac{n!}{k!m!}$,
\begin{align} \nonumber\exp(x+y)&=\sum_{n=k+m} \frac{1}{n!}\frac{n!}{k!m!}x^ky^m\\ &=\exp(x)\cdot\exp(y) \end{align}

2. Would this result still hold if x and y are complex numbers? What about matrices?
Note that in Binomial theorem,
$$\sum _{k=0}^{n}{n \choose k}x^{n-k}y^{k}=\sum _{k=0}^{n}{n \choose k}x^{k}y^{n-k}$$
does NOT work for matrices, so it does NOT hold for matrices.

Reference:

- YouTube: What is Euler’s formula actually saying? | Lockdown math ep. 4

- Proving the Exponential Functional Equation exp(x+y)=exp(x)exp(y)