Deriving properties of the exp function from definition Define exponential function $\exp$ as follows:
$$\exp:\Bbb{R}\rightarrow(0,\infty)$$
$$\begin{align}i) \ \forall x,y\in \Bbb{R}:\exp(x+y)=\exp(x)\cdot\exp(y)\end{align}$$
$$\begin{align}ii) \ \forall x\in \Bbb{R}:\exp(x)\ge x+1 \end{align}$$
Book says these 4 properties can be derived using this definition.
$$(1) \ \exp \text{is increasing on } \Bbb{R} $$
$$(2) \ \exp(0)=1$$
$$(3) \ \forall x\in \Bbb{R}: \forall n \in \Bbb{Z}: \exp(nx)=(\exp(x))^n$$
$$(4) \ \forall x\in \Bbb{R},x>0:\exp(x)>1$$
The only thing i managed to do myself was to prove $(3)$ by mathematical induction for $n\in\Bbb{N}$. I also tried to prove the monotonicity directly from definition, but can't seem to find a starting point. Would you please give me hint on where to start with proofs of these? Another thing I am interested in is a proof of uniqueness of this definition. How do i prove, assuming only these two definitions $i),ii)$ that $\exp$ function defined like this is unique?
EDIT:
For the proof of $(2)$ i have following idea:
$$\exp(0)=\exp{(1+(-1))}=\exp(1)\cdot\exp(-1)$$
and by $(3)$ (not proven yet for negative integers) we can continue
$$\exp(1)\cdot\exp(-1)=\frac{\exp(1)}{\exp(1)}=1$$
EDIT 2: What remains is to show that $(3)$ holds for negative values of $n$.
 A: By your condition (ii), $\exp(x) > 1$ for all $x > 0$. Hence if $y > x$, we have
$$\exp(y) = \exp(x + (y - x)) = \exp(x) \cdot \exp(y - x) > \exp(x) \cdot 1$$
giving monotonicity. The fact that $\exp(0) = 1$ follows from setting $x = y = 0$ and knowing that $\exp(0) > 0$.

As for uniqueness, the standard technique is to take two candidates and compare them. For example, if $\exp_2$ is another function satisfying the conditions, study $\exp - \exp_2$ and make sure it's the zero function. Or something more natural here is to take $\exp / \exp_2$ and show it's identically one.
A: I have never come across this definition of the exponential. Where did you find this? Apart from that, let me help you: 
Note that we can take $x=y=0$ in i) to find $\mbox{exp}(0)=1$ (it cannot be zero because of ii) as you can see). Also, if we have (4) then we also have (1) by using (i) again. In fact, $\mbox{exp}$ is then strictly increasing. So it remains to prove (4).
If there is an $x>0$ such that $\mbox{exp}(x)=1$, then by (3), which we already proved, we obtain a set of arbitrarily large real numbers $nx$ for which $\mbox{exp}(nx)=1$. This must contradict ii).
As for uniqueness, I think you need something a little more sophisticated. Can you prove that $\mbox{exp}$ is differentiable with derivative equal to itself, using the properties you established? If this is the case, you know that its derivative at $0$ must equal one because otherwise property (ii) cannot hold. Observe that this property ii) is actually very restrictive, because for example if we only use (1)-(4) as assumptions, any function of the form $x \mapsto a^x$ would fit the bill for positive $a$. 
Edit: as an elaboration, if we know that the function is differentiable with derivative equal to itself, then the quotient of any two functions satisfying i) and ii) must have zero derivative by application of the quotient rule, so that it is identically $1$.  
