In the entry exponential function one defines for real variable $x$ the real exponential function $\exp{x}$ , i.e. $e^x$ , as the sum of power series: $$e^x \;=\; \sum_{k=0}^\infty\frac{x^k}{k!}$$ The series form implies immediately that the real exponential function attains only positive values when $x \geqq 0$ . Also for $-1 \leqq x < 0$ the positiveness is easy to see by grouping the series terms pairwise.

In order to study the sign of $e^x$ for arbitrary real $x$ , we may multiply the series of $e^x$ and $e^{-x}$ using Abel's multiplication rule for series. We obtain $$e^xe^{-x} \;=\; \sum_{n=0}^\infty\frac{x^n}{n!}\!\sum_{k=0}^\infty(-1)^k\frac{x^k}{k!} \;=\; \sum_{n=0}^\infty\!\sum_{j=0}^n(-1)^j\frac{x^n}{j!(n\!-\!j)!} \;=\; \sum_{n=0}^\infty\frac{x^n}{n!}\!\sum_{j=0}^n\!{n\choose j}(-1)^j \;=\; \sum_{n=0}^\infty\frac{x^n}{n!}\cdot0^n.$$ The last sum equals 1. So, if $-x > 0$ , then $e^{-x} > 0$ , whence $e^x$ must be positive.

Let us now consider arbitrary complex value $z = x\!+\!iy$ where $x$ and $y$ are real. Using the addition formula of complex exponential function and the Euler relation, we can write $$e^z \;=\; e^{x+iy} \;=\; e^xe^{iy} \;=\; e^x(\cos{y}+i\sin{y}).$$ From this we see that the absolute value of $e^z$ is $e^x$ , which we above have proved to be positive. Accordingly, we may write the

Theorem. The complex exponential function never vanishes.