TITLE: Alternate Proof for $e^x \ge x+1$
QUESTION [6 upvotes]: This is just a standard problem from my high school's calculus text, but my proof seems sort of off. This is it:
Let $f(x) = e^x$. The tangent line of $f(x)$ at $x=0$ is $g(x)=x+1$. Since $f''(x_0) \gt 0$ for all $x_0 \in \mathbb R$,the tangent line $g(x) \le f(x)$ for all $x$. Q.E.D.
I didn't show all of the work, but is there something wrong here? It is pretty short. Anyways, I would like to see some alternate proofs because I tried to think of another but when my brain said, "meh", I realized that there are probably tons of ways to prove this. So just any valid proof would be cool with me. I don't care if they're crazy. It's even cooler when they're crazy. But in particular the simplest proof would be appreciated.

REPLY [3 votes]: When $x>=0$, $e^x>=1+x$.
When $x <-1$, $\;1+x < 0$, so $e^x>=1+x$
When $ -1\leq x \lt 0$, then lets take $y = -x$, so, 
$$e^x=e^{-y}=(1-y)+y^2(1/2!-y/3!+\cdots>=(1-y)+y^2((1/2!-1/3!)+y^2(1/4!-1/5!)+\cdots)\geq (1-y)\geq(1+x)$$
[Remember, $(1/n!-1/(n+1)!>0$]