Analogously, here are several ways to define me:
I am the citizen of the US with social security number [XYZ]. This is of primary interest to the government.
I am the oldest son of [my mother's name]. This is of primary interest to my family.
I am the instructor of [particular course meeting at particular days/times] at [university]. This is of primary interest to students in that class.
I am the author of [a particular master's thesis]. This is (maybe) of primary interest to my thesis advisor.
Of the above list, which is "the right definition" of me?
As you can see, I am related to the world in a multitude of very specific ways. Though some are quite different in their nature, they all determine me uniquely, with different people and institutions thinking of me primarily in different ways.
Similarly, the constant $e$ is related to various pieces of mathematics in many different, but specific, ways. The definition used may vary depending on what role $e$ is fulfilling in a particular context, but they all uniquely determine the same constant and are all important for their own reasons.
There are a few things I say over and over again on math.SE, and one of them is that the cleanest definition of the exponential function (on either $\mathbb{R}$ or $\mathbb{C}$, or more generally even) is that it's the unique function $f : \mathbb{C} \to \mathbb{C}$ (or $\mathbb{R} \to \mathbb{R}$) satisfying
- $f(0) = 1$, and
- $f'(x) = f(x)$.
Note that this definition makes no explicit reference to $e$. Every other property of the exponential function falls easily out of this definition together with the existence and uniqueness theorems for solutions to ODEs. For example, by the chain rule
$$\frac{d}{dz} \exp(z + w) = \exp(z + w)$$
hence $\exp(z + w)$ is also a solution to the above ODE but with initial condition $\exp(w)$. But so is $\exp(z) \exp(w)$. Hence the two are equal by the uniqueness theorem.
Similarly we get continuity at every point and the usual power series expansion. The limit
$$\exp(z) = \lim_{n \to \infty} \left( 1 + \frac{z}{n} \right)^n$$
then falls out of applying the Euler method with step size $\frac{z}{n}$ to approximate solutions to this ODE. (It can also be formally justified by differentiating with respect to $z$ but this requires some thought about exchanging the derivative and the limit.)
This allows us to give a clean definition of $e$ as just being the value $\exp(1)$ (another thing I say over and over again on math.SE is that $e$ is not important, $\exp(z)$ is important and $e$ just happens to be its value at $z = 1$), and a clean definition of $\pi$: with $\exp(z)$ defined as above, $\pi$ is the smallest positive real such that $\exp(2 \pi i) = 1$, or in other words it's half the period of $\exp(it)$. Note that by the chain rule
$$\frac{d}{dt} \exp(it) = i \exp(it)$$
so $\exp(it)$ is a solution to the ODE $f(0) = 1, f'(t) = i f(t)$ for a function $f : \mathbb{R} \to \mathbb{C}$. But $\cos t + i \sin t$ is also such a solution. So by the uniqueness theorem we recover Euler's formula
$$\exp(it) = \cos t + i \sin t.$$
This requires that we know what the trigonometric functions are in advance, but we can actually invent them this way instead. Additivity gives $\exp(it) \exp(-it) = \exp(0) = 1$, but we also have
$$\frac{d}{dt} \exp(it) \overline{\exp(it)} = 0$$
from which it follows that $\exp(-it) = \overline{\exp(it)}$ and that $\| \exp(it) \| = 1$ is a constant. So $\exp(it) = c(t) + i s(t)$ satisfies
$$c(t)^2 + s(t)^2 = 1$$
$$c(-t) = c(t), s(-t) = - s(t)$$
$$c(t_1 + t_2) = c(t_1) c(t_2) - s(t_1) s(t_2)$$
$$s(t_1 + t_2) = c(t_1) s(t_2) + s(t_1) c(t_2)$$
and we're well on our way to rediscovering trigonometry. These identities can be used to show that $\exp(it)$ is periodic by showing that it not only lies on the unit circle but travels on it with constant velocity (this basically follows from additivity).
The same uniqueness idea applied to the trigonometric functions tells us that $(\cos t, \sin t)$ is the unique pair of functions satisfying
- $c(0) = 1, s(0) = 0$, and
- $c'(t) = -s(t), s'(t) = c(t).$
Every other trigonometric identity is a consequence of these. This one may be a little less intuitive but it says that the vector $\left[ \begin{array}{cc} c'(t) \\ s'(t) \end{array} \right]$ is a $90^{\circ}$ rotation of, and in particular orthogonal to, the vector $\left[ \begin{array}{cc} c(t) \\ s(t) \end{array} \right]$, which e.g. after differentiating a second time, exactly describes a particle under the influence of a constant centripetal force.
Best Answer
The following are equivalent definitions for $\exp(x)$. \begin{align} 1. & f(x) = \sum_{k=0}^{\infty} \dfrac{x^k}{k!}\\ 2. & \dfrac{d f(x)}{dx} = f(x) \text{ with } f(0) = 1\\ 3. & f(x) = \lim_{n \to \infty} \left(1+\dfrac{x}n\right)^n\\ 4. & f(x+y) = f(x) \cdot f(y) \text{ with }f(x) >0 \text{ being continuous at one point and } f(1) = e \end{align} If you start with any one, you can derive/prove the others.
EDIT
The important thing is that you can start with anyone and derive the others as property. If a statement $A$ implies a statement $B$ and vice-versa, both are equivalent statements. We may, hence, use any one of them as a definition.
For instance, if you choose $(1)$ to define $\exp(x)$ as $\exp(x) = \displaystyle \sum_{k=0}^{\infty} \dfrac{x^k}{k!}$, the rest from $(2)$ to $(4)$ become properties.