How come such different methods result in the same number, $e$?

This identity, in isolation, is certainly confusing. It becomes less so once you know that it is just a special case of a more general identity

$$\sum_{n \ge 0} \frac{x^n}{n!} = \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n.$$

Why does this hold? Well, let's call the function on the left $e^x$. Then by inspection, $\frac{d}{dx} e^x = e^x$. General theory tells you that the solution to $y' = y$ is unique provided you fix the value of $y(0)$ (and this is also physically intuitive), and here $e^0 = 1$ by inspection, so that tells you that $e^x$ is the unique function satisfying $e^0 = 1$ and $\frac{d}{dx} e^x = e^x$.

Chandru1's deleted answer reminded me that the proof of this is actually completely elementary. Suppose $f$ is a function which satisfies $f'(x) = f(x)$. Then

$$\frac{d}{dx} \left( e^{-x} f(x) \right) = e^{-x} f'(x) - e^{-x} f(x) = 0$$

hence $e^{-x} f(x) = C_f$ for some constant $C_f$. Substituting $f(x) = e^x$ gives $e^{-x} e^x = C$, and substituting $x = 0$ gives $e^{-x} e^x = 1$. Now it follows that $f(x) = C_f e^x$.

Why does the function on the RHS also satisfy this property? Let's call this function $\exp(x)$. Then

$$\exp(cx) = \lim_{n \to \infty} \left( 1 + \frac{cx}{n} \right)^n = \lim_{m \to \infty} \left( 1 + \frac{x}{m} \right)^{mc} = \exp(x)^c$$

where $m = \frac{n}{c}$. Setting $c = 1 + \frac{y}{x}$, this gives

$$\exp(x + y) = \exp(x)^{1 + \frac{y}{x}} = \exp(x) \exp(x)^{ \frac{y}{x} } = \exp(x) \exp(y)$$

from which it follows that

$$\frac{d}{dx} \exp(x) = \lim_{h \to 0} \frac{\exp(x) \exp(h) - \exp(x)}{h} = \exp(x) \lim_{h \to 0} \frac{\exp(h) - 1}{h}$$

so it follows that $\exp(x)$ satisfies $\exp(0) = 1$ and $\frac{d}{dx} \exp(x) = \exp(x)$, which means it must equal $e^x$ - at least, as soon as we know that

$$\lim_{h \to 0} \frac{\exp(h) - 1}{h} = 1.$$

But

$$\exp(h) = \lim_{n \to \infty} \left( 1 + \frac{h}{n} \right)^n = \lim_{n \to \infty} \left( 1 + {n \choose 1} \frac{h}{n} + {n \choose 2} \frac{h^2}{n^2} + ... \right) = \lim_{n \to \infty} \left( 1 + h + R(h, n) \right)$$

where $|R(h, n)| \le \sum_{k \ge 2} |h|^k = O(|h|^2)$ by the ratio test, from which the conclusion follows.

That last step, by the way, is an echo of the standard proof, where one expands

$$\left( 1 + \frac{x}{n} \right)^n = \sum_{k=0}^n {n \choose k} \frac{x^k}{n^k}$$

and uses the fact that $\frac{1}{n^k} {n \choose k} \to \frac{1}{k!}$ as $n \to \infty$. While this is fairly hands-on, in my opinion it obscures an underlying principle behind this problem, which is that $e^x$ is a very special function and all of the equivalent ways of defining it are based on one of its special properties.

One really nice way to interpret the above identity is that it is precisely what you get by applying Euler's method to the ODE $y' = y$ with initial conditions $y(0) = 1$.

Another way of viewing $e^{x}$, is if suppose $f : \mathbb{R} \to \mathbb{R}$ is a differentiable function which satisfies $f'(x) = f(x)$ for all $x \in \mathbb{R}$, then $f(x) = e^{x} + C$. To see this we have $f'(x) = f(x) \Longrightarrow e^{-x}f'(x) = e^{-x} f(x) \Longrightarrow e^{x} \cdot [f'(x) - f(x)]= 0 \Longrightarrow \frac{\rm d}{\rm dx } \Bigl[e^{-x}f(x)\Bigr]=0$ which says that $e^{-x} f(x) = C$. Hence $f(x) = Ce^{x}$.