Define $$\ln (x) = \int^{x}_{1}\frac{1}{t}$$
Assume I have proven that $\ln x$ is one-to-one and therefore has an inverse $\exp (x)$.
Define $e$ as:
$\ln e = 1$
Now, if you have no other notion of exponentials, or logarithms, how could define what $e^x$ means and show that its the inverse of $\ln x$?
You are allowed to assume the logarithmic product and quotient property.
Thanks for the help.
On
Writing $f$ for $\ln$ and $g$ for its inverse, you can show easily that $g$ is infinitely differentiable and that $g^{(n)}(0) =1$. This gives you the Taylor series $g(x) = \sum_{n=0}^{\infty}x^n/n!$. After that, everything follows from the classical analysis of $g$ that is performed in every elementary real variables text (see Rudin's Real & Complex Analysis, for example).
As I recall, the introductory "Chapter 0" of that text is a marvel of succinct mathematics that fully constructs the exponential function from scratch. It's really a pleasure to read and I'm always awed at his insight every time I read it.
Full disclosure: this is essentially a rewrite of MPW's answer.
Defining $e^x$ as $\sum\limits_{k=0}^{\infty} {x^k\over{k!}}$ makes sense, in a way it's the most fundamental/general definition because it can be applied to any system for which addition, multiplication and scaling are defined. Reals, complex numbers, quaternions, matrices, etc.
Now, from calculus we have this result: $$\frac{d}{dx} \left[ f^{-1}(x) \right] = {1\over{f'(f^{-1}(x))}}$$
By fundamental theorem of calculus we obtain $\ln'(x)$ as $1\over{x}$, hence: $$\frac{d}{dx} \left[ \ln^{-1}(x) \right] = \ln^{-1} (x)$$
It follows (formally by induction) that the $n$th derivative of $\ln^{-1}(x)$ is $\ln^{-1}(x)$ and hence the $n$th derivative at $x = 0$ is $\ln^{-1}(0) = 1$.
Thus we get the Taylor series: $$ \ln^{-1}(x) = \sum\limits_{k=0}^{\infty} {x^k\over{k!}} = e^x$$