proof of derivative of an exponential function

Note that we have $$\frac{b^h-1}{h}=\frac{e^{h\log(b)}-1}{h} \tag 1$$

Now, in THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequality

$$e^x\ge 1+x \tag 2$$

From $(2)$ (along with the property $e^xe^{-x}=1$) it is easy to see that for $x<1$

$$e^x\le \frac{1}{1-x} \tag 3$$

Using $(2)$ and $(3)$, we can bound $(1)$ as

$$\log(b) \le \frac{e^{h\log(b)}-1}{h}\le \frac{\log(b)}{1-h\log(b)}$$

whereupon application of the squeeze theorem yields the coveted limit

$$\lim_{h\to 0}\frac{b^h-1}{h}=\log(b)$$

And we are done!