You talk about "deriving" the value of e, but, in fact, in most texts the value of e is defined just as you give it:
From the derivative calcluation, we can show that the derivative of a^x for any positive a is
a^x \lim_{h\to 0}\frac{a^h- 1}{h}
We define "e" to be the value of a that makes that coefficient 1:
\lim_{h\to 0}\frac{e^h- 1}{h}= 1
We can roughly (and it can be made precise by "\delta- \epsilon arguents") say that, for h close to 0, we have, approximately,
\frac{e^h- 1}{h}~ 1
e^h- 1~ h
e^n~ 1+ h
e~ (1+ h)^h
and so arrive at the definition e= \lim_{h\to 0} (1+ h)^h which is equivalent to
e= \lim_{n\to\infty} (1+ \frac{1}{n})^{n}
You will also see that definition of "e" in situations that have nothing to do with derivatives. If you have a loan, or investment, or bank account, where the interest rate is r and the interest is compounded n times a year, after one year the return (or amount owed if a loan) will be the original amount times (1+ r/n)^n. The limit, as n goes to infinity, of that is \lim_{n\to\infty}(1+ r/n)^n= e^r.
As for specific values of e, we can see that if a= 2, taking h= .001 gives (a^h- 1)/h= .6939< 1 while taking a= 3, h= .001, (a^h- 1)/h= 1.099> 1 so the value of a= e, giving a limit of 1 must be between 2 and 3. Taking a= 2.5, h= .001, (a^h- 1)/h=0.917< 1 so e must be between 2.5 and 3. Taking a= 2.75, h= .001, (a^h- 1)/h=1.012 so e must be between 2.5 and 2.75. Continuing in that way we can get e as accurately as we wish.
I will also point out that many new Calculus texts first define ln(x) by
ln(x)= \int_1^x \frac{1}{t} dt
All of the properties of ln(x) can be shown from that, including that it is invertible. We can then define exp(x) to be the inverse function. It can then be shown that, for all x, exp(x)= (exp(1))^x so, defining e= exp(1), we have exp(x)= e^x. That is the same as saying that ln(e)= 1 so that e can be evauated numerically by repeated numerical integrations of 1/x from 1 to different values of a, each time getting the integral closer to 1.