# Proof

(related to Proposition: Natural Logarithm)

Let $$n\in\mathbb N$$. From the definition of the exponential function, it follows that

$\exp(n)=1+n+\frac {n^2}2+\ldots \ge 1+n.\quad \quad ( * )$

Due to the reciprocity of exponential function, we get with $$( * )$$ the approximation

$\exp(-n)=\frac 1{\exp(n)}\le \frac 1{1+n}.\quad\quad ( * * )$

Since the exponential function is strictly monotonically increasing and is continuous, we can conclude from the lemma about invertible functions that the it is invertible on closed real intervals $$\left[-n,n\right]$$ for all $$n\in\mathbb N$$. Note that $$\bigcup_{n\in\mathbb N}[-n,n]=\mathbb R$$ and $$\bigcup_{n\in\mathbb N}\left[\frac 1{n+1},n+1\right]=\mathbb R_+^*$$, where $$\mathbb R_+^*$$ denotes the set of all positive real numbers. Therefore, it follows that the natural logarithm
$\ln:\mathbb R_{+}^*\to\mathbb R$ is the inverse function of the exponential function and that it is also continuous, strictly monotonically increasing.

Thank you to the contributors under CC BY-SA 4.0!

Github:

### References

#### Bibliography

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