Proof that the number $e$ exists based on the following definition My book defines the number $e$ as follows (Simmons' Calculus With Analytic Geometry, 2nd edition, pg. 265):

$e$ is the number for which $\lim_{h\to0}\frac{e^h-1}{h}=1$

However, it does not provide a proof that such a number exists. How can we prove that? Thanks very much in advance.

As you have stated that your book defines $e$ as the number for which
$$\lim_\limits{h\to 0}\frac{e^h-1}{h}=1$$
I must say that, the definition is not mathematically sound and circular in nature and also seems confusing as to whether it uses the exponential function to define $e$.
However, what I think it intended to say is that

If, for some number $x$, we have $$\lim_\limits{h\to
 0}\frac{x^h-1}{h}=1$$ then we can say that $x=e$.

This might look similar to your book's definition, but this one is comparatively clear and mathematically sound; still it is circular and I hereby prove it.
This definition can be rewritten as $$\lim_\limits{h\to 0}\frac{x^h-1}{h}=1$$
$$\Rightarrow \lim_\limits{h\to 0}\frac{x^{0+h}-x^0}{h}=1$$
$$\Rightarrow \lim_\limits{h\to 0}\frac{f(0+h)-f(0)}{h}=1 \,\,\,\,\,\, \text{where} \,\,\,f(p)=x^p$$
$$\Rightarrow \frac{d}{dp}(x^p)=1 \,\,\,\,\,\, \text{at} \,\,\,\, p=0$$
$$\Rightarrow x^p \ln x=1 \,\,\,\,\,\, \text{at} \,\,\,\, p=0$$
$$\Rightarrow  \ln x=1 $$
$$\Rightarrow  x=e $$
So this definition implies that you have already defined natural logarithm without defining $e$, which is quite baseless and illogical.
