Using the fact that the number e is specifically chosen to satisfy the differential equation $\frac{d(e^x)}{dx} = e^x$ it's pretty easy to see that $e = \lim_{n\to\infty}(1+\frac{1}{n})^n$ (or at least, I could derive this.)
It's pretty well known that when you end up differentiating an exponential function $f(x) = n^x$, you get $f'(x)=n^x\ln(n)$
If $f(x) = f'(x)$, then $\ln(n)=1$, and $n=e$, which is consistent.
Using the standard definition of the derivative, for $f(x) = n^x$, you end up getting
$$f'(x)=n^x\lim_{h\to0}\frac{x^h-1}{h}$$
This implies that
$$\ln(x) = \lim_{h\to0}\frac{x^h-1}{h}$$
but how do you prove this (which will also end up proving that differential rule stated above, and will also prove that $e^{\lim_{h\to0}\frac{x^h-1}{h}}=x$)
Best Answer
$$f(x)=n^x\qquad n>0\\ f'(x)=\lim_{h\to 0}\frac{n^{x+h}-n^x}{h}\\ f'(x)=n^x\lim_{h\to 0}\frac{n^h-1}{h}\\ L=\lim_{h\to 0}\frac{n^h-1}{h}\\ L=\ln n\lim_{h\to 0}\frac{e^{h\ln n}-1}{h\ln n}$$ Now substitute: $$t:=h\ln n\\ t\to 0\quad\text{as}\quad h\to 0\\ L=\ln n\lim_{t\to 0}\frac{e^t-1}{t}$$ Notice by the definition of $e$: $$\frac{d(e^x)}{dx}=e^x\lim_{t\to 0}\frac{e^t-1}{t}=e^x\\ \lim_{t\to 0}\frac{e^t-1}{t}=1\\ L=\ln n\\ f'(x)=n^x\ln n$$