Show that $(\exp(x)\cdot \exp(x')=\exp(x+x')$ and $ \exp(x)>\exp(0),x>0)\Rightarrow \exp$ is strictly increasing $exp(x)=\sum_{n=0}^{\infty}\frac{1}{n!}x^n$
My calculations so far:
Assume 
$\exists _{x,x'\in\mathbb{R}}:$
$x<x'$ and $\exp(x)>\exp(x')$
$x<x'\Rightarrow x'-x>0 \Rightarrow \exp(x-x')>\exp(0)$
Can I somehow invoke a contradiction with this?
 A: First let $x=x'=0$ therefore $$\exp(0)\cdot\exp(0)=\exp(0)$$which yields to $$\exp(0)=0\text{ or }1$$ if $\exp(0)=0$ then we have $$\exp(x){=\exp(x+0)\\=\exp(0)\cdot\exp(x)\\=0\quad,\quad \forall x\in \Bbb R}$$which is non-sense. Therefore $\exp(0)=1$. Now let $x'>x$. Then there exists $\epsilon>0$ for which $x'=x+\epsilon$. This leads to$$\exp(x'){=\exp(x+\epsilon)\\=\exp(x)\exp(\epsilon)\\>\exp(x)\exp(0)\\=\exp(x)}$$therefore$$\exp(x')>\exp(x)$$which completes our proof.
A: Let $y>x \leftrightarrow y-x>0$. Then
$exp(y)=exp(y-x+x)=exp(y-x) exp(x)> exp(0) exp(x) =exp(x)$.
In the second and third step you use the two assumptions.
A: In general, for any $f(\cdot)$ with $f(x+y)=f(x)f(y)$ where $f(y)>f(0)=1$ for every $y>0$, you can immediately prove that $f(\cdot)$ is strictly increasing. 
For this, I start by proving $f(x)>0$ for every $x$. To see this, use $f(x+y)=f(x)f(y)$, take $y=-x$ for an $x>0$ and get $f(x)f(-x)=f(0)>0$. Since $f(x)>0$, we deduce $f(-x)>0$.
Now, fix an $x\in\mathbb{R}$, and let $z>x$. Then, $z=x+\epsilon$ for an $\epsilon>0$. This gives, $f(z)=f(x)f(\epsilon)>f(x)$, as $f(x)>0$ and $f(\epsilon)>1$. Done.
