# Exponential series/Real/Elementary properties/Fact/Proof

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Proof

(1) follows directly from the definition.
(2) follows from

${\displaystyle {}\exp x\cdot \exp {\left(-x\right)}=\exp {\left(x-x\right)}=\exp 0=1\,}$

using fact.
(3) follows for ${\displaystyle {}n\in \mathbb {N} }$ from fact by induction, and from that it follows with the help of (2) also for negative ${\displaystyle {}n}$.
(4). Nonnegativity follows from

${\displaystyle {}\exp x=\exp {\left({\frac {x}{2}}+{\frac {x}{2}}\right)}=\exp {\frac {x}{2}}\cdot \exp {\frac {x}{2}}={\left(\exp {\frac {x}{2}}\right)}^{2}\geq 0\,.}$

(5). For real ${\displaystyle {}x}$ we have ${\displaystyle {}\exp x\cdot \exp {\left(-x\right)}=1}$, so that because of (4), one factor must be ${\displaystyle {}\geq 1}$ and the other factor must be ${\displaystyle {}\leq 1}$. For ${\displaystyle {}x>0}$, we have

${\displaystyle {}\exp x=\sum _{n=0}^{\infty }{\frac {1}{n!}}x^{n}=1+x+{\frac {1}{2}}x^{2}+\ldots >1\,}$

as only positive numbers are added.
(6). For real ${\displaystyle {}y>x}$, we have ${\displaystyle {}y-x>0}$, and therefore, because of (5) ${\displaystyle {}\exp {\left(y-x\right)}>1}$, hence

${\displaystyle {}\exp y=\exp {\left(y-x+x\right)}=\exp {\left(y-x\right)}\cdot \exp x>\exp x\,.}$