$\exp(x)$ is usually defined in three different ways:
1) By its Taylor series: $\exp(x)=\sum_{k=0}^{\infty} \frac{x^k}{k!}$
2) By its derivative: $\exp(x)'=\exp(x)$
3) By the limit $\exp(x)=\lim_{N \rightarrow \infty} \left(1+\frac{x}{N} \right)^N$
In textbooks they mostly use the 3rd way, because $\exp(x)$ is introduced before the derivative.
But it just occured to me - we don't need this much information to define this function. There is a very simple definition which allows to recover all the other properties:
4)
$$\lim_{x \rightarrow 0} \exp(x) = \lim_{x \rightarrow 0} (1+x)$$
Now, it's just two terms of Taylor expansion - it doesn't seem like much. Yet we can prove the following properties:
a) It follows that $\lim_{x \rightarrow 0} \exp(a x) = \lim_{x \rightarrow 0} \exp^a(x)$
$$\lim_{x \rightarrow 0} \exp^a(x) = \lim_{x \rightarrow 0} (1+x)^a = \lim_{x \rightarrow 0} (1+a x) = \lim_{x \rightarrow 0} \exp(a x)$$
b) Using this property we recover the 3rd definition
$$ \lim_{N \rightarrow \infty} \left( 1+\frac{x}{N} \right)^N=\lim_{N \rightarrow \infty} \left( \exp \left(\frac{x}{N}\right) \right)^N=\lim_{N \rightarrow \infty} \left( \exp^{\frac{1}{N}}(x) \right)^N=\exp(x) $$
Does this work, or did I make a mistake somewhere?
Edit
I was obviously wrong - there is an infinite number of functions with the property 4. So the definition needs to be:
4) For $x \rightarrow 0$ $$ \exp(x) \approx 1+x$$
For any $x$
$$\exp^a(x) = \exp(a x)$$
$$\lim_{x\to 0} (1+x)$$ is simply equal to $1$, so the equation you wrote is equal to the equation $$\lim_{x\to0} \exp(x) = 1$$
which is most certainly not enough to characterise the exponential function, for example $\cos(x)$.
As for what you proved in points (a) and (b), I don't even know what you were intending to prove so I cannot comment on that.