Alternate Proof for $e^x \ge x+1$ 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.
 A: Sure, your proof is pretty short, but it is only short for someone who already did some analysis. For example, you used the following theorems (which is true, of course) in your proof:

If $f$ is convex on $\mathbb R$, then for any tangent function $g$, we
  know that $g(x)<f(x)$ for all $x$.

and

If $f$ is twice differentiable on $\mathbb R$ and $f''(x)>0$ for all
  $x$, then $f$ is convex on $\mathbb R$

You also used the definition of convex functions (implicitly) to prove your statement.
In short, I would say that your proof is short and elegant, but not simple.

Another way to prove your point is to simply see what the minimum value of $e^x-x-1$ is.
A: I don't know how you have defined the exponential function, but you could use the fact that it has the following series expansion $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} +\cdots$$
A: My favourite proof is letting $f(x)=e^x-x-1$. $f(0)=0$ and it's easy to show that it has a minimum at $x=0$. 

However you do it, you are somehow going to go through derivatives or calclulus, explicitly or implicitly.
EDIT:
A completely alternative proof: Using the taylor series ($e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} +...$), it is easy to show the inequality for $x>-1$ (grouping the terms together by twos, the even ones are greater than the odd ones, the even ones being positive and the odd ones negative). For $x \leq -1$, $LHS>0$ while $RHS \leq 0$, so we are done.
A: 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, $(1/n!-1/(n+1)!>0$]
A: OK, here's yet another way of looking at it:  consider the function
$\rho(x) = e^{-x}(1 + x); \tag{1}$
we have
$\rho(0) = 1; \tag{2}$
also
$\rho'(x) = -e^{-x}(1 + x) + e^{-x} = e^{-x}(1 - (1 + x)) = -x e^{-x}. \tag{3}$
Thus
$\rho'(x) < 0 \; \; \text {for} \; \; x > 0, \tag{4}$
and
$\rho'(x) > 0 \; \; \text{for} \; \; x < 0. \tag {5}$
Sooooo, for $x > 0$,
$\rho(x) - 1 = \rho(x)- \rho(0) = \int_0^x \rho'(t) dt < 0, \tag{6}$
whence
$\rho(x) < 1 \tag{7}$
or, multiplying by $e^x$,
$1 + x < e^x. \tag{8}$
For $x < 0$,
$1 - \rho(x) = \rho(0) - \rho(x) = \int_x^0 \rho'(t) dt > 0, \tag{9}$
yielding again
$\rho(x) < 1, \tag{10}$
and again
$e^x < 1 + x. \tag{11}$
QED.
We note that that this argument uses just a tad of calculus, i.e. the fundamental theorem of same and the fact that $\rho(x)$ is differentiable; oh yeah, and that integrals of positive smooth functions are positive etc.  Also, we have shown that
the inequalities are strict if $x \ne 0$.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
