Q:

derivative of exponential function equals f'(0)

In my James Stewart Calculus book (Page 178) I'm completely confused by what this part means:
Notice that the limit is the value of the derivative of $f$ at 0, that is,
$$\lim_{h \to 0} \frac{b^h - 1}{h} = f'(0)$$
Therefore we have shown that if the exponential function $f(x) = b^x$ is differentiable at 0, then it is differentiable everywhere and $$f'(x) = f'(0)*b^x$$
How does that limit equal f'(0) and how does that does tell us the exponential function is differentiable everywhere? 

A:

If $f(x) = b^x$ then
$$f'(x) = \lim_{h \to 0} \frac{b^{x+h}-b^x}{h}=\lim_{h \to 0} b^x\frac{b^{h}-b^0}{h}=b^x\lim_{h \to 0} \frac{b^{h}-1}{h}=b^x f'(0).$$
Consequently, if $f'(0)$ exists then so does $f'(x) = b^x f'(0)$.

