Definition of e, and proof of existence. So, I was wondering what is the proof of the existence of $e$. I want to define it as the number such that for which the derivative of $(b^x)'=b^x$. Now, I have to show it exists. So, I have to show that there is a number $b$ such that $\lim{\frac{b^h-1}{h}}=1$ but how would I prove that. Any nice and easy proofs? Maybe it is easier to use another definition of $e$ and show that that $e$ also fulfils this definition?
 A: To do this the "logarithm first" way, write
$$ \frac{b^h-1}{h} = \int_1^b x^{h-1} \, dx. $$
We now have to use a result about interchanging limits and integrals: a simple one is that since $x^{h-1}$ are continuous functions, converging uniformly to the continuous function $x^{-1}$ on $[0,b]$ as $h \to 0$, the limit and integral can be interchanged, so
$$ \lim_{h \to 0} \int_1^b x^{h-1} \, dx = \int_1^b \frac{dx}{x}. $$
The latter, normally called the logarithm of $b$, $\log{b}$, is an increasing function of $b$ that starts from $0$ and is unbounded above, so the intermediate value theorem implies there is a value of $b$ for which it has the value $1$.
A: As suggested by Jack d'Aurizio via comment, you have to proceed step by step:


*

*Define $b^{x} =\lim_{n\to\infty} b^{x_{n}} $ where $x_{n} $ is a sequence of rationals converging to $x$. You need to show that this definition makes sense: the limit exists and is independent of the specific sequence $x_{n} $. 

*Establish that the limit $\lim_{x\to 0}(b^{x}-1)/x$ exists for all $b>0$ and defines a function of $b$ say $L(b) $. 

*Show that $L(b) $ is a strictly increasing function of $b$ and maps interval $(0,\infty)$ to whole of $\mathbb{R} $. And hence $L(b) =1$ for a unique number $b>0$ which is traditionally denoted by $e$. Your problem is solved now but do continue further. 

*Show that $L(ab) =L(a) +L(b) $ and $\dfrac{d}{db} L(b) =\dfrac{1}{b}$.


All the above steps are carried out in almost the same order in my blog post. The function $L$ is traditionally denoted by $\log$. By the way the approach outlined above is intuitive but difficult. There are easier approaches available to develop the theory of exponential and logarithmic functions which are described here and here. 
A: Expanding on the comment of Francesco, let's start out with your hypothetical number $b$, such that the function $b^x$ differentiates to itself. In addition, let's blindly hope that this function is given by a (absolutely convergent) power series:
$$ b^x = \sum_{n = 0}^\infty a_n x^n $$
For now we just hope that this works, and see what the coefficients would have to be. Since $b^0 = 1$, we have that $a_0 = 1$. Taking derivatives on both sides gives that $(b^x)' = b^x = a_1 + 2 a_2 x + 3 a_3 x^2 + \cdots$, and so $b^0 = 1$ gives $a_1 = 1$. Keep differentiating and evaluating at $x = 0$ to find that in general, $a_n = \frac{1}{n!}$. So we have a candidate function:
$$ b^x = \sum_{n = 0}^\infty \frac{1}{n!}x^n = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$
You can apply some standard analysis tests (root test, ratio test, etc) to show that this function is convergent on the whole of $\mathbb{R}$. (This is essentially the "hard" step in the existence here). Furthermore, we have that
$$ \frac{b^h - 1}{h} = \frac{h + \frac{h^2}{2!} + \frac{h^3}{3!} + \cdots}{h} = 1 + \frac{h}{2!} + \frac{h}{3!} + \cdots$$
Which has a limit of $1$ as $h \to 0$. Finally, you can recover the number $e$ as a value of this function: $e = b^1$.
As an interesting fact, notice here that the only properties of the function $b^x$ actually used were that it was its own derivative, and $b^0 = 1$. This was enough to completely determine the exponential function.
