calculus - Proving that $lim_{hto 0 } frac{b^{h}-1}{h} = ln{b}$

Is there a formal proof of this fact without using L'Hôpital's rule? I was thinking about using a proof
of this fact:

$$\left.\frac{d(e^{x})}{dx}\right|_{x=x_0} = e^{x_0}\lim_{h\to 0} \frac{e^{h}-1}{h} = e^{x_0}\cdot 1=e^{x_0}$$
that I have to help prove:
$$\lim_{h\to 0} \frac{b^{h}-1}{h} = \ln{b}$$
Is there a succinct proof of this limit? How does one prove this rigorously?

Answer

It is easy to see that $$\lim_{x\to 0}\frac{\log_a(1+x)}{x}=\log_a e$$ Now set $y=a^x-1$. So $a^x=1+y$ and then $$x=\log_a(1+y)$$ Clearly, when $x\to0$; $y\to 0$ so $$\lim_{x\to 0}\frac{a^x-1}{x}=\lim_{y\to 0}\frac{y}{\log_a(1+y)}=\log_e a$$