How can we prove that there is a number $a$ such that $\lim_{h\to 0}\frac{a^{h} -1}{h}=1$? One definition of $e$ that I am fond of is that it is the number $a$ such that
$$
\lim_{h\to 0}\frac{a^{h} -1}{h}=1
$$
The reason for this is that it cuts to the heart of the special property of all exponential functions. If we have $f(x)=a^x$, then
\begin{align}
f'(x)&=\lim_{\Delta x \to 0}\frac{a^{x+\Delta x} -a^x}{\Delta x} \\
&=\lim_{\Delta x \to 0}a^x\frac{a^{\Delta x} - 1}{\Delta x} \\
&=a^x\lim_{\Delta x \to 0}\frac{a^{\Delta x} - 1}{\Delta x}
\end{align}
However, presumably there is a caveat to this approach. We need to show that a number like $e$ exists in the first place! In other words, we need to show that $g(a)=\lim_{\Delta x \to 0}\frac{a^{\Delta x} -1}{\Delta x}$ takes the value of $1$ for some value of $a$. How might we do this?

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 $h \to 0-$.
