Why does $(1 + x)^{1/x} $ approach $e$ as $x$ approaches $0$? For smaller and smaller values of x in the expression $(1 + x)^{1/x} $, its value seems to approach the constant $e$,

$(1 + 1E-10)^{1E10} = 2.71828182832313114395...   $
$(1 + 1E-20)^{1E20} = 2.718281828459045235347$

but how can it be proven that:
$$\lim_{x\to 0}(1 + x)^{1/x}  = e$$
I was first introduced to e with this definition
$$\frac{d(e^x)}{dx}=e^x$$
Hints:

With L'Hospital show that $ \frac{\ln(1+x)}{x} \to 1$ as $x \to 0.$

We have $(1+x)^{1/x}= e^{\frac{\ln(1+x)}{x}}.$
Use binomial expansion:
$$f(x)=(1+x)^{1/x}=1+x\frac{1}{x}+\frac{\frac{1}{x}(\frac{1}{x}-1)}{2!}x^2+\frac{\frac{1}{x}(\frac{1}{x}-1)(\frac{1}{x}-2)}{3!}x^3+\frac{\frac{1}{x}(\frac{1}{x}-1)(\frac{1}{x}-2)(\frac{1}{x}-3)}{4!}x^4+...$$
$$f(x)=1+1+\frac{(1-x)}{2!}+\frac{(1-x)(1-2x)}{3!}+\frac{(1-x)(1-2x)(1-3x)}{4!}+...$$
$$\lim_{x \to 0} f(x)=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...= e$$