Why the differentiation of $e^x$ is $e^x?$ $$\frac{d}{dx} e^x=e^x$$
Please explain simply as I haven't studied the first principle of differentiation yet, but I know the basics of differentiation.

Here is the computation of the derivative of the general $y= a^x$ where a can be any positive number.
$y(x+ h)- y(x)= a^{x+ h}- a^x= (a^x)(a^h)- a^x= a^x(a^h- 1)$
So $\frac{y(x+h)- y(x)}{h}= a^x\left(\frac{a^h- 1}{h}\right)$
$\frac{da^x}{dx}= \lim_{h\to 0}a^x\left(\frac{a^h- 1}{h}\right)$
$= a^x\lim_{h\to 0}\left(\frac{a^h- 1}{h}\right)$
So the derivative of $a^x$ is $a^x$ times some constant, $\lim_{h\to 0}\frac{a^h- 1}{h}$.
It is easy to see that, if a= 1, since $a^h= 1$ for all x, that limit is 0 and if a= 3, since $3^{0.001}= 1.001099$,  approximately, $\frac{1.01099- 1}{0.001}= 1.099$, there is some a, between 1 and 3, such that $\
lim_{h\to 0} \frac{a^{x+h}- a^x}{h}= 1$.
We define "e" to be that value of a so that the derivative  of $e^x$ is 1 times $e^x$.
