Proof $e^x = \exp(x)$? 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.
 A: 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.
A: 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: $$ \ln^{-1}(x) = \sum\limits_{k=0}^{\infty} {x^k\over{k!}} = e^x$$
A: Preliminaries:
First, you don't need to assume the product/quotient rule, as they follow from the definition:
$$\ln (xy) = \int^{xy}_{1}\frac{1}{t}=\int^{x}_{1}\frac{1}{t} +\int^{xy}_{x}\frac{1}{t}= \int^{x}_{1}\frac{1}{t} dt +\int^{xy}_{x}\frac{x}{xt}dt$$
Now, if you make the substitution $u=xt$ in the second integral, you get:
$$\ln (xy) =  \int^{x}_{1}\frac{1}{t} dt +\int^{y}_{1}\frac{1}{u}du=\ln(x)+\ln(y)$$
The quotiont rule for the logarithm follows from here, or can be proven the same way.
Second, it follows from the FTC that $\ln'(x)=\frac{1}{x}$.
Finally, the definition of the logarithm makes sense for $x>0$, and $\ln'(x)=\frac{1}{x} >0$ on its domain, which implies that $\ln(x)$ is one to one. It therefore has an inverse.
It can also be proved from the definition that the range of $\ln$ is $\mathbb R$. Therefore it has an inverse $\exp(x) : \mathbb R \to (0, \infty)$, and all these properties follow from your definition, nothing needs to be assumed.
The proof that $\exp(x)=e^x$
Since $\ln(xy)=\ln(x)+\ln(y)$ its inverse function satisfy the functional equation 
$$\exp(x+y)=\exp(x) \exp(y)$$
From here [exactly as in the Cauchy equation], one gets by induction that 
$$\exp(nx)=[\exp(x)]^n$$
for all positive integers, and then for all integers.
From here it also follows that 
$$\exp(\frac{p}{q})=[\exp(1)]^{\frac{p}{q}} \,.$$
As you defined $e=\exp(1)$, this shows that 
$$\exp(r)=e^r \, \forall r \in \mathbb Q \,.$$
Finally, since $e >1$ (which follows from the definition of the logarithm), the function $e^x$ is increasing [note that since you defined $e=\exp(1)$ you should not use the derivative here, as you don't know if your didn't get a different $e$].
Moreover, it follows immediately from the above definition of $\ln(x)$ that $\ln(x)$ is a strictly increasing function, and therefore, so is its inverse.
Now, your claim follows immediately from the following lemma applied for f(x)=e^x, g(x)=\exp(x)$:
Lemma Let $f(x), g(x)$ be two increasing functions. If $f(x)$ is continuous and $f(x)=g(x) \forall x \in \mathbb Q$ then $f=g$.
The proof is easy, for each $x \in \mathbb R$ pick two sequences of rationals, one which is increasing to $x$, and the other which is decreasing to $x$ and use monotony + continuity of $f$.
Bonus To show that your definition of $e$ is the standard one, note that since $\ln'(x)=\frac{1}{x}$ the formula for the derivative of the inverse function yields
$$[e^x]'=e^x$$
