How to calculate $\lim \limits_{h \to 0}{\frac{a^h-1}{h}}$? As the title says, I would like to prove for $f(x) = a^x$ there is always some constant c such that $f'=cf$. Is calculating the limit the right approach to solve this problem? Also, how to show there is only one solution when $c=1$? (the $e^x$)

Assuming $a>0$, you need to use the definition of real exponentiation:
$$a^h=e^{h\ln a}=\sum_{n=0}^\infty\frac{(h\ln a)^n}{n!}$$
Then
$$\lim_{h\to0}\frac{a^h-1}h=\lim_{h\to 0}\sum_{n=1}^\infty h^{n-1}\frac{(\ln a)^n}{n!}=\ln a$$
