# Proof

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

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

are absolutely convergent series for all $$x,y\in\mathbb R$$. 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 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