Statement
The derivative of $ e^x $ is $ e^x $:
$$ \tfrac{d}{dx}(e^x) = e^x $$
The definition of e
Take the definition of $ e $.
$$ e = \lim_{n \to \infin}\bigg(\left(1 + \tfrac{1}{n}\right)^n\bigg) $$
When substituting $ n = \frac{1}{h} $, $ h $ goes to $ 0 $ as $ n $ goes to $ \infin $.
$$ e = \lim_{h \to 0}\bigg((1 + h)^\frac{1}{h}\bigg) $$
Later in the proof this definition of $ e $ is used.
Proof
Define function $ f $.
$$ f(x) = e^x $$
Take the definition of the derivative and substitute.
$$ f'(x) = \lim_{h \to 0}\left(\frac{f(x + h) - f(x)}{h}\right) $$
$$ f'(x) = \lim_{h \to 0}\left(\frac{e^{x + h} - e^x}{h}\right) $$
By the rule of adding exponents, write $ e^{x + h} = e^x * e^h $ and factor out $ e^x $.
$$ f'(x) = \lim_{h \to 0}\left(\frac{e^x * e^h - e^x}{h}\right) $$
$$ f'(x) = \lim_{h \to 0}\left(\frac{e^x(e^h - 1)}{h}\right) $$
Since $ e^x $ does not include the limit variable, $ h $, bring it out of the limit.
$$ f'(x) = \lim_{h \to 0}\left(e^x * \frac{e^h - 1}{h}\right) $$
$$ f'(x) = e^x * \lim_{h \to 0}\left(\frac{e^h - 1}{h}\right) $$
Now substitute $ e $ in the limit with what is defined above. Since that definition uses $ h \to 0 $ and this limit does too, it is not required to write the limit again.
$$ f'(x) = e^x * \lim_{h \to 0}\left(\frac{\big((1 + h)^\frac{1}{h}\big)^h - 1}{h}\right) $$
Multiply the exponents and cancel like terms.
$$ f'(x) = e^x * \lim_{h \to 0}\left(\frac{1 + h - 1}{h}\right) $$
$$ f'(x) = e^x * \lim_{h \to 0}\left(\frac{h}{h}\right) $$
Finally, since $ \frac{h}{h} $ is $ 1 $, the limit disappears.
$$ f'(x) = e^x * \lim_{h \to 0}\big(1\big) $$
$$ f'(x) = e^x $$