> [!NOTE] Lemma > For all $x\in \mathbb{R},$ $1+x\leq \exp(x)$where $\exp$ denotes the [[Real Exponential Function|real exponential function]]. **Visualisation**: ![[Inequalities for the exponential.png|400]] **Proof**: This proof assumes the definition of $\exp$ as the [[Real Exponential Function as Power Series|power series]]: $\exp(x)=\sum_{n=0}^{\infty} \frac{x^{n}}{n!}.$ If $x \geq 0$ then $\exp(x)=1+x+ \frac{x^{2}}{2}+\dots \geq 1+ x$If $x\leq -1$ then $1+x\leq 0$ whereas $\exp(x)>0$ so $\exp(x)\geq 1+x.$ Now suppose $-1<x<0.$ Then by [[Upper Bound for Real Exponential Function for all x<1]], $\exp(-x)\leq \frac{1}{1+x}$ which implies $1+x\leq e^{x}.$