Exponential Inequality

$\text { Consider a function } f(x)=e^x-x-1 \\ f^{'}(x)=e^x-1 \\ f^{'}(x)>0 \hspace{5mm} \forall \hspace{5mm} x>0 \\ f(x) \text{ is an increasing function } \\ \implies f(x) > f(0) \\ e^x-x-1 > 0 \\ e^x > x+1$

$e^x = \sum_{i=0}^{\infty} \frac{x^i}{i!} = 1+x+\sum_{i=2}^{\infty} \frac{x^i}{i!}$

The terms in the last summation are all positive whenever $x>0$ . Therefore,

$\boxed{e^x > 1+x}$ when $x>0$