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How to prove that the exponential function is strictly positive ?

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$$e^x = \sum \frac {x^n} {n!} $$

Starting with this definition, it is obvious that $e^x > 0$ when $x \geq 0$. To show that $e^{-x} > 0$ it is enough to show that $e^{-x} \cdot e^x = 1$.

$$\sum_{n=0} \frac {(-x)^n} {n!} \cdot \sum_{m=0} \frac {x^m} {m!} = \sum_{m,n} \frac {(-x)^nx^m} {n!m!} = \sum _k \sum_{m+n = k} \frac{((-x)+x)^{n+m}}{(n+m)!} = 1$$

Here the second equality is due to $(a+b)^n = \sum_k \binom n k a^{n-k}b^k$ and the last one is due to $0^0 = 1$.

The exponential function $\exp$ is the unique(!) solution of the differential equation $$\tag1 y’=y,\qquad y(0)=1.$$

However, if $\exp(a)\ne0$, then $z\mapsto \frac1{\exp(a)}\exp(z+a)$ is also a solution. By uniqueness, we conclude $\exp(z)=\frac1{\exp(a)}\exp(a+z)$, provided $\exp(a)\ne 0$.

Especially, $\exp(2z)=\exp(z)^2$ if $\exp(z)\ne 0$.

Let $z\in\mathbb C$ be arbitrary. Then from $2^{-n}z\to 0$ and $\exp(0)=1$ and continuity, we conclude that $\exp(2^{-n}z)\ne 0$ for some $n\in\mathbb N$. Then by repeated squaring $$\tag2\exp(z)=(\cdots(\exp(2^{-n}z\underbrace{)^2)^2)\cdots)^2}_n\ne 0.$$

We conclude that the complex exponential has no zeroes.

In the real case, $(2)$ shows that $\exp(z)>0$ because squaring a nonzero real number makes it positive.

*Remark:* If you start with a different definition of the exponentail, for example via the well-known power series, you can readily show that $(1)$ holds.

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