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.
That's a subtle and interesting question. A precise answer will depend on what your book has previously assumed or proved about exponentiation. Modulo that, I'll attempt an outline of an argument.
The very statement of that definition assumes you know about raising arbitrary positive numbers to arbitrary exponents. Then you'd prove that for fixed $b$ the limit $$ A(b) = \lim_{h \rightarrow 0} \frac{b^h -1}{h} $$ exists (along the way proving that $f(x) = b^x$ is differentiable and proportional to its derivative).
Then show $A$ is monotonically increasing. Clearly $A(2) < 1$ and $A(3) > 1$ so there's a number $e$ between $2$ and $3$ for which $A(e) = 1$. Looking back, you identify $A(x) = \ln(x)$.
I doubt that your book has done the work required to make this rigorous.
Nonetheless, I think it's a good intuitive way to get at $e$. It's what you need to describe the function that's its own derivative. That's what makes $e$ "natural" and why it appears in many problems in growth and decay.
I am of the opinion that that definition is very very wrong, because $e$ is there defined by using $e$, you can see $e$ on the right-hand side in the expression $\lim_{h\to0}\frac{e^h-1}{h}=1$.
So, to prove that such an $e$ exists it is necessary to define it in some other way, perhaps as a limit of some sequence (as usual), and then go on to define exponential function and then use the properties of that function to prove that limit.
The way $e$ is defined there in that book is very very circular, at least in my opinion.
Or, as Randall wrote, you could show that there is one and only one number $a$ for which $\lim_{h\to0}\frac{a^h-1}{h}=1$ and then call that $a$ by the name $e$, but the things in your book seem to be different.
Define $f(x) = \lim_{h \to 0} \frac{x^h - 1}{h}$. Clearly $f(1) = 0$ and (handwaving the proof that the derivative and limit commute) $f'(x) = \lim_{h \to 0} \frac{h x^{h-1}}{h} = 1/x$, so $f(x) = \int_1^x \frac{d\xi}{\xi}$. You can prove that this integral increases without bound at $x \to \infty$ (even without knowing that it's a logarithm) by comparing it to the harmonic series, so $f$ is monotonically increasing and unbounded. The existence and uniqueness of $f^{-1}(1)$ follow.
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.
