How can we prove that there is a number $a$ such that $\lim_{h\to 0}\frac{a^{h} -1}{h}=1$?

Independently, we can prove that there exists a number $e$ such that $e = \lim_{n \to \infty} (1+1/n)^n$ along with the property that for all $n \in \mathbb{N}$,

$$\left(1 + \frac{1}{n} \right)^n < e < \left(1 + \frac{1}{n} \right)^{n+1}$$

It follows that

$$1 < n(e^{1/n}-1) < n\left[\left(1 + \frac{1}{n} \right)^{1/n}\left( 1+ \frac{1}{n}\right)-1 \right] \leqslant 1 + \frac{1}{n} + \frac{1}{n^2},$$

where the right-hand inequality is obtained using Bernoulli's inequality $(1 + 1/n)^{1/n} \leqslant 1 + 1/n^2$.

By the squeeze theorem we get

$$\tag{*}\lim_{n \to \infty}\frac{e^{1/n}-1}{\frac{1}{n}} = 1$$

From here it is not difficult to show that

$$\tag{**} \lim_{h \to 0+}\frac{e^{h}-1}{h} = 1$$

Taking $n = \lfloor1/h\rfloor$ when $h > 0$, we have $n \leqslant 1/h < n+1$ and

$$\frac{n}{n+1}(n+1)(e^{1/n+1} - 1)= n(e^{1/n+1} - 1 ) \leqslant \frac{e^h -1 }{h} \leqslant (n+1)(e^{1/n} -1) = \frac{n+1}{n}n(e^{1/n}-1)$$

Since $n \to \infty$ if and only if $h \to 0$ we obtain (**) by the squeeze theorem using the previous result (*).

With some more work we can show that the limit $1$ is also attained as $h \to 0-$.