Prove inequality $e^{x}>1+x$ Prove that $e^{x}>1+x$ when $x\neq0$. I can't make for all $x$. But as $e^{x}>0$ for all $x$, it's obvious that $e^{x}>1+x$ when $x\le-1$. It's where I got got stuck
 A: Take $f(x) = e^x -(1+x)$. Since $f'(x) = e^x -1 >0$ iff $x>0$, it means that $f$ is increasing on $(0,+\infty)$ and decreasing on $(-\infty,0)$. Moreover $f(0)=0$. Then it follows that $f>0$ if $x \neq 0$.
A: The graph of $e^x$ is concave up everywhere, and thus lies above any tangent line to that graph. Since $y = 1+x$ is the tangent line at $x = 0$, the result follows.
A: Hint:$$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}=1+x+\sum_{n=2}^{\infty}\frac{x^n}{n!}$$
A: $f(x)=e^x$ is a convex function because $f''(x)>0$, thus
$$f'(x)>f(x_1)+f'(x_1)(x-x_1)$$
set $x_1=0$
A: $f(x)=e^x-(1+x) \Rightarrow  f'(x)=e^x-1 \Rightarrow f''(x)=e^x>0$
So the minimum point is given by $f'(x)=0 \Rightarrow x=0$, then $f(x)>0$ if $x \ne 0$
A: 
In THIS ANSWER, I showed that the sequence $\displaystyle e_n(x)=\left(1+\frac xn\right)^n$ monotonically increases for $x>-n$. 

Therefore, for $n>1$ and $x>-n$
$$e_n(x)\ge e_1(x)=1+x$$
Letting $n\to \infty$, we find the coveted inequality
$$e^x=\lim_{n\to \infty}\left(1+\frac xn\right)^n\ge 1+x$$
for all $x$.  And we are done!
A: These sorts of questions are inherently problematic because you have not specified the definition of $e^x$ you're working with. I usually define $\log$ as an integral.  It's easy to show that it's a strictly increasing $C^{\infty}$ bijection $\mathbb{R_+} \to \mathbb{R}$. We define $e^x$ as its inverse. It is also $C^{\infty}$, strictly increasing and bijective $\mathbb{R} \to \mathbb{R_+}$. 
After that, it's not too hard to prove the standard properties of the function, such as $(e^x)' = e^x$. 
Note
$$e^x= 1 + \int_{0}^x e^t dt \ \ (1) $$
We've specified $e^t$ is positive, so $e^x \geq 1$ since the integral in $(1)$ must be non-negative. Referring to $(1)$ in a recursive manner, we can see that since $e^x \geq 1$, $\int_{0}^x e^t dt \geq \int_{0}^x \ dt = x$ so that $e^x \geq x+1$. 
This argument can be generalized to show that $$e^x \geq \sum_{i=0}^n \frac{x^i}{i!}$$ for all $n \in \mathbb{N}$. 
