Show that $e^x > 1 + x + x^2/2! + \cdots + x^n/n!$ for $n \geq 0$, $x > 0$ by induction Show that if $n \geq 0$ and $x>0$, then 
$$ e^x > 1 + x + \frac{x^2}{2!} + \dots + \frac{x^n}{n!}.$$
Not sure where to get started with this induction proof.

Base case: $e^x > 1$ for $x > 0$.
Induction: suppose we are given $k$ such that for all $x > 0$
$$
e^x > 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^k}{k!}
$$
Change variables to $t$ and integrate both sides:
$$
\int_0^x e^t \; dt > \int_0^x \left( 1 + t + \frac{t^2}{2!} + \cdots + \frac{t^k}{k!} \right) \; dt
$$
$$
e^x - 1 > x + \frac{x^2}{2!} + \cdots + \frac{x^{k+1}}{(k+1)!}
$$
$$
e^x > 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^{k+1}}{(k+1)!}
$$
Note that if $f > g$ on $(a,b)$ and both are integrable, then $\int_a^b f > \int_a^b g$ (not just $\ge$).
