Definition of exponential function Suppose we have no knowledge of the value of $e$
You are interested in characterizing the functions $F = \{f \in \mathbb{R^{\mathbb{R}}}: f' = f\}$
It isn't too hard to show that if $f \in F$, such that $f > 0$ somewhere, then $f > 0$ everywhere and $f$ is a monotonically increasing function. Additionally, $f \in F \implies af \in F, a \in \mathbb{R}$, by the chain rule. Thus, if $\exists f \in F$ with $f>0$, then $\exists g\in F$ with $g(0) = 1$.
Q: How can we show $g(x) = \sum_{k=0}^{\infty}\frac{x^k}{k!}$
 A: If you have some general theory of power series, you can show that $\sum_n x^n/n!$ converges everywhere to a function $\exp(x)$ that satisfies $dy/dx=y$.
Since the differential equation is linear, therefore,
$$ g(x) = f(x)- \exp(x) $$
satisfies the same differential equation, and is $0$ for $x=0$.
You have already proved that this prevents $g$ from being positive anywhere. Likewise, this prevents $g$ from being negative anywhere, so $g$ is identically zero, and $f(x)=\exp(x)$.

To justify the notation $e^x$:
Once you know (by the above argument) that the differential equation has a unique solution, you can prove that $\exp(x+y)=\exp(x)\exp(y)$ -- namely, for fixed $y$, the function $x\mapsto \frac{1}{\exp(y)}\exp(x+y)$ satisfies $f'=f$ and $f(0)=1$, and it must therefore equal $\exp(x)$.
We can now, by induction and a bit of busywork to handle signs, prove
$$ \exp(x)^n = \exp(nx) $$
for arbitrary $n\in\mathbb Z$, and then
$$ \exp(x)^{n/m} = \exp(\tfrac nm x) $$
Finally, setting $x$ to $1$ and defining $e=\exp(1)$ yields
$$ e^q = \exp(q) $$
for all $q\in\mathbb Q$.
If we have already defined irrational powers of positive reals by requiring continuity, this ensures $e^x = \exp(x)$ for all $x\in\mathbb R$ (because $\exp$ is differentiable and therefore continuous). Otherwise, define $b^x=\exp(x\exp^{-1}(b))$ and we see that this agrees with the algebraic definition of rational powers and is continuous in both $b$ and $x$ for $b>0$.
A: hint
we have $$\forall x\in \Bbb R \;\;g'(x)=g (x) \;\;g (0)=1$$ 
By Cauchy-Lipschitz, there is only one solution.
we check easily that
$$x\mapsto 1+x+\frac {x^2}{2!}+... $$
has an infinite radius of convergence and it satisfies the conditions above.
