To show that $e^x > 1+x$ for any $x\ne 0$ $$e^x>1+x$$ 
is  what I want to show.
So let's define a function:
$$h\left(x\right)=e^x-x-1$$
and investigate its derivative: 
$$h'\left(x\right)=e^x-1$$.
Easy to see that at $x=0$ it has a critical point and it is a minimum, and therefore for any other value that isn't $0$ we'll get a value that is bigger then the minimum $f(0)=0$, in other words, for each $x\ne 0$, $h\left(x\right)>0$, meaning $e^x-x-1>0$ meaning that:
$$e^x>x+1$$
 A: It's a local minimum. In order to establish that it's a global minimum, you'll need to do more than take the derivative and apply the first or second derivative test.
One theorem you can use is Rolle's theorem (or the more general version, the mean value theorem). It states that if you have a function $f : [a, b] \rightarrow \mathbb{R}$ that is continuous, as well as differentiable on $(a, b)$, with $f(a) = f(b)$, then $f'(c) = 0$ for some $c \in (a, b)$.
In particular, you've established that $h(0) = 0$, and presumably that $h'(x)$ exists everywhere and is equal to $0$ exactly when $x = 0$. Then, if there's any other point $x$ where $h(x) = 0$, by Rolle's theorem, there must exist some point $c$ strictly between $0$ and $x$ with $h'(c) = 0$. But, $h'(c)$ cannot be $0$, since $c \neq 0$. This is a contradiction, so the minimum is global.
A: .....Your proof is correct.....
A: Looks good to me. It is worth noting that one definition of $e^x$ is $e^x = \sum_{0}^{\infty} \frac{x^n}{n!}$. When looking at this definition, is is clear that the proposition is true. Since this is an infinite sum of polynomial terms, the exponential function would dominate every polynomial for large enough $x$.
A: Since $h(0)=0$, we have $h(x)=\int_0^x h'(t)\,dt=\int_0^x e^t-1\,dt$.
Now, $t \mapsto e^t$ is increasing and so $e^t>e^0=1$ for $t>0$ and  $e^t<e^0=1$ for $t<0$.
This implies that $h(x)=\int_0^x e^t-1\,dt >0$ because the integrand is positive when $x>0$ and negative when $x<0$. In the latter case, $\int_0^x e^t-1\,dt = \int_x^0 -(e^t-1)\,dt >0$.
A: I think our OP's proof has the right idea, especially when taken in light of the comments.   Here's a way to work it which avoids explicit use of minima/maxima, and hence avoids any pitfalls inherent in such methods:
One can also look at the function
$\alpha(x) = e^{-x}(1 + x); \tag{1}$
we have
$\alpha(0) = 1 \tag{2}$
and
$\alpha'(x) = -e^{-x}(1 + x) + e^{-x}  = -xe^{-x}; \tag{3}$
we see that
$x > 0 \Rightarrow \alpha'(x) < 0 \tag{4}$
and
$x < 0 \Rightarrow \alpha'(x) > 0; \tag{5}$
thus when $x > 0$,
$\alpha(x) - 1 = \alpha(x) - \alpha(0)$
$= \int_0^x \alpha'(s) ds < 0, \tag{6}$
so
$\alpha(x) < 1, \tag{7}$
whence
$1 + x < e^x; \tag{7}$
also, for $x < 0$,
$1 - \alpha(x) = \alpha(0) - \alpha(x)$
$= \int_x^0 \alpha'(s) ds > 0, \tag{8}$
thus again
$\alpha(x) < 1, \tag{9}$
showing that
$1 + x < e^x \tag{10}$
in this case as well.
A: The second derivative of $e^x$ is $e^x,$ which is everywhere positive. Any function with strictly positive second derivative lies above any of its tangent lines except at the points of tangency. The line tangent to $y=e^x$ at $(0,1)$ is $y=x+1.$ Therefore $e^x>1+x, x\ne 0.$
