First, let's prove that :
$$(1)\space\space e = \lim_{n \to \infty} \bigg (1+\frac{1}{n} \bigg)^n$$
For that, we need to know :
$$(2)\space\space\lim_{x \to 0} \frac{ln(1 + x)}{x}$$
Let's say : $\forall x > -1,\space f(x) = ln(1 + x)$
$f(0) = ln(1) = 0$, so :
$$\frac{ln(1+x)}{x} = \frac{f(x) - f(0)}{x} = \frac{f(0+x) - f(0)}{(0 + x) - 0}$$
By using the definition of the derivative,
$$\lim_{x \to 0} \frac{ln(1 + x)}{x} = \lim_{x \to 0} \frac{f(0+x) - f(0)}{(0 + x) - 0} = f'(0)$$
$$\forall x > -1,\space f'(x) = \frac{1}{1+x}\space And\space f'(0) = 1$$
Finally,
$$\lim_{x \to 0} \frac{ln(1 + x)}{x} = 1$$
Now (2) is computed, let's prove (1) :
Let's say : $\forall x \neq 0, \space y=\frac{1}{x}$
So :
$$x = \frac{1}{y}$$
$x \to 0 \implies y \to \infty$
$$\lim_{x \to 0} \frac{ln(1 + x)}{x} = \lim_{y \to \infty} \frac{ln(1 + \frac{1}{y})}{\frac{1}{y}} = \lim_{y \to \infty} y \space ln\bigg(1 + \frac{1}{y}\bigg)=\lim_{y \to \infty} ln\bigg(1+\frac{1}{y}\bigg)^y$$
$$\lim_{y \to \infty} ln\bigg(1+\frac{1}{y}\bigg)^y = \lim_{x \to 0} \frac{ln(1 + x)}{x} = 1$$
$$(3)\space\space\lim_{y \to \infty} ln\bigg(1+\frac{1}{y}\bigg)^y = 1$$
Let's say exp the inverse function of ln :
$$exp(1) = \lim_{y \to \infty} \bigg(1+\frac{1}{y}\bigg)^y$$
Let's say $e = exp(1)$, and replace y by n, we finally prove :
$$(1)\space\space e = \lim_{n \to \infty} \bigg (1+\frac{1}{n} \bigg)^n$$
(1) could also be written :
$$(1)\space\space e = \lim_{f(n) \to \infty} \bigg (1+\frac{1}{f(n)} \bigg)^{f(n)}$$
Let's say : $\forall x \neq 0$, $f(n) = \frac{n}{x}$
Let's separate the proof in 3 cases : (a) $x>0$, (b) $x<0$ and (c) $x=0$
Case (a) : $x>0$ so $f(n) \to \infty \implies n \to \infty$
$$e = \lim_{n \to \infty} \bigg (1+\frac{1}{\frac{n}{x}} \bigg)^{\frac{n}{x}} = \lim_{n \to \infty} \bigg (1+\frac{x}{n} \bigg)^{\frac{n}{x}}$$
By raising at power x :
$$\forall x>0, \space e^x = \lim_{n \to \infty} \bigg (1+\frac{x}{n} \bigg)^{n}$$
Case (b) : $x<0$ so $f(n) \to \infty \implies n \to -\infty$
$$e = \lim_{n \to -\infty} \bigg (1+\frac{1}{\frac{n}{x}} \bigg)^{\frac{n}{x}} = \lim_{n \to -\infty} \bigg (1+\frac{x}{n} \bigg)^{\frac{n}{x}}$$
Let's say $m = -n$
$n \to -\infty \implies m \to \infty$
$$e = \lim_{m \to \infty} \bigg (1+-\frac{1}{\frac{m}{x}} \bigg)^{-\frac{m}{x}} = \lim_{m \to \infty} \bigg (1-\frac{x}{m} \bigg)^{\frac{m}{-x}}$$
By raising at power -x :
$$e^{-x} = \lim_{m \to \infty} \bigg (1-\frac{x}{m} \bigg)^{m}$$
By replacing -x by x and m by n :
$$\forall x < 0, \space e^{x} = \lim_{n \to \infty} \bigg (1+\frac{x}{n} \bigg)^{n}$$
(1) and (2) implies :
$$\forall x \neq 0, \space e^{x} = \lim_{n \to \infty} \bigg (1+\frac{x}{n} \bigg)^{n}$$
Case (c) : $e \neq 0 \implies e^0 = 1$
Let's see if the formula valid for $x \neq 0$ is also valid for $x=0$ :
$$\space e^{0} = \lim_{n \to \infty} \bigg (1+\frac{0}{n} \bigg)^{n} = \lim_{n \to \infty} 1^n = 1$$
We just proved:
$$\forall x \in \mathbb{R}, \space e^{x} = \lim_{n \to \infty} \bigg (1+\frac{x}{n} \bigg)^{n}$$