You really should be looking for definitions of the exponential function $e^x$, not definitions of $e$. Here are the most important ones that come to mind:
- $e^x$ is the unique function $f(x)$ satisfying $f'(x) = f(x)$ and $f(0) = 1$.
- $e^x$ is the inverse of the function $\displaystyle \ln x = \int_1^x \frac{dt}{t}$.
- $e^x$ is the power series $\displaystyle \sum_{n \ge 0} \frac{x^n}{n!}$.
- $e^x$ is the limit $\displaystyle \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n$.
I say this because once you start thinking about $e^x$, which is by far the more fundamental object, and not $e$, the relationship between the definitions becomes much more transparent. Here are short sketches of proofs that the definitions above are all equivalent:
$1 \Leftrightarrow 2$: if $\frac{d}{dx} g(x) = \frac{1}{x}$ then
$$\frac{d}{dx} g^{-1}(x) = \frac{1}{g'(g^{-1}(x))} = g^{-1}(x).$$
Conversely, if $\frac{d}{dx} g(x) = g(x)$ then
$$\frac{d}{dx} g^{-1}(x) = \frac{1}{g'(g^{-1}(x))} = \frac{1}{g(g^{-1}(x))} = \frac{1}{x}.$$
$1 \Leftrightarrow 3$: if $f'(x) = f(x)$ then $f^{(n)}(x) = f(x)$ for all $n$, hence $f^{(n)}(0) = f(0) = 1$, so the Taylor series of $f(x)$ has all coefficients equal to $1$. Conversely, the function with that Taylor series is its own derivative and satisfies $f'(0) = 1$ using the fact that power series are term-by-term differentiable inside their interval of convergence.
$1 \Leftrightarrow 4$: $\displaystyle \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n$ is the result of using Euler's method to compute $f(x)$, where $f$ satisfies $f'(t) = f(t)$ and $f(0) = 1$, as the step size $n$ goes to $\infty$.
(My personal opinion is that the first definition is the most fundamental one; in general uniqueness statements are very powerful. For example, there is a very short proof using the first definition, which I invite you to find, that $e^{x + y} = e^x e^y$.)