# Proof

Let $$x,y\in\mathbb R$$ be complex numbers. By definition of the complex exponential function,

$\exp(x)=\sum_{n=0}^\infty\frac{x^n}{n!}\quad\text{ and }\quad\exp(y)=\sum_{n=0}^\infty\frac{y^n}{n!}$

are absolutely convergent complex series for all $$x,y\in\mathbb C$$. Therefore, their Cauchy product. $\sum_{n=0}^\infty c_n=\left(\sum_{n=0}^\infty\frac{x^n}{n{!}}\right)\cdot\left(\sum_{n=0}^\infty\frac{y^n}{n{!}}\right)$ is also an absolutely convergent complex series. Moreover, applying the binomial theorem, we get $c_n:=\sum_{k=0}^n\frac{x^{n-k}}{(n-k)! }\cdot\frac{y^k}{k! }=\frac1{n!}\sum_{k=0}^n\binom nk x^{n-k}y^k=\frac{(x+y)^n}{n! }.$ It follows $\exp(x+y)=\sum_{n=0}^\infty\frac{(x+y)^n}{n! }=\left(\sum_{n=0}^\infty\frac{x^n}{n{ ! }}\right)\cdot\left(\sum_{n=0}^\infty\frac{y^n}{n{! }}\right)=\exp(x)\cdot \exp(y).$

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### References

#### Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer VerĂ¤nderlichen", Vieweg Studium, 1983