If the natural logarithm is defined by the integral, how to show that exp is the inverse? If we take $\ln(x) = \int_1^x \frac{1}{s}ds$ as the definition of the natural logarithm, how can we show, that the exponential function $e^x$ is its inverse (With respect to the proper domains/codomains of course).
Also it's an undergraduate question, it appeared surprisingly hard to me, so I think I miss something here... 
 A: Since you're allowing any sound definition of $e^x$ except the one being 'the inverse of $\ln(x)$', I'm going to (conveniently) pick this one

Definition: we define the exponential function as the solution of the following differential equation, whose existence is guaranteed by Picard's theorem:

$$
f'(x) = f(x)  \ , \ f(0) = 1
$$

We will note $e^x := f(x)$.

You can show this is equivalent to the power series definition with minimal notions of complex analysis.
Now, by the fundamental theorem of calculus, we know that $\ln(x)$ is differentiable with derivative $\frac{1}{x}$ and so is $e^x$ (by definition) with derivative $e^x$. Now,
$$
(\ln(e^x))' = \frac{e^x}{e^x} = 1
$$
so $\ln(e^x)= x + c$ for some constant $c$, but evalutating at $x = 0$ gives that $c = 0$. Now, for the other composition, 
$$
(e^{\ln(x)})' = \frac{e^{\ln(x)}}{x}
$$
so $g(x) = e^{\ln(x)}$ statisfies $g'(x) = \frac{g(x)}{x}$, $g(1) = 1$, and therefore $g(x) = x$, simply because there is (again, by Picard) only one solution and $id$ satisfies the conditions. 
Since both compositions give the identity, $\ln(x)$ and $e^x$ are inverses in the corresponding domains.
A: Let's use the definition $e^x=\lim_{n\to\infty}\left(1+{x\over n}\right)^n$. 
Begin by noting that the substitutions $t^n=u=1+v$ give us
$$\int_1^{\left(1+{x\over n}\right)^n}{dt\over t}=n\int_1^{1+{x\over n}}{du\over u}=n\int_0^{x/n}{dv\over1+v}=n\int_0^{x/n}\left(1-{v\over1+v}\right)dv=x-n\int_0^{x/n}{v\over1+v}dv$$
where we assume $|x/n|\lt1$ so that we're not integrating over $t=0$ in case $x$ is negative, and
$$n\left|\int_0^{x/n}{v\over1+v}dv \right|\le n\left|\int_0^{|x/n|}{v\over1-|x/n|}dv \right|={n|x/n|^2\over2(1-|x/n|)}={|x|^2\over2n(1-|x/n|)}\to0$$
as $n\to\infty$. Now we also have
$$\int_1^{e^x}{dt\over t}=\int_1^{\left(1+{x\over n}\right)^n}{dt\over t}+\int_{\left(1+{x\over n}\right)^n}^{e^x}{dt\over t}$$
(still assuming $|x|\lt n$ to stay away from $t=0$) and
$$\left|\int_{\left(1+{x\over n}\right)^n}^{e^x}{dt\over t} \right|\le\max\left\{e^{-x},\left(1+{x\over n}\right)^{-n}\right\}\left|e^x-\left(1+{x\over n}\right)^n \right|\to0$$
as $n\to\infty$. Thus
$$\ln(e^x)=\int_1^{e^x}{dt\over t}=\lim_{n\to\infty}\left(\int_1^{\left(1+{x\over n}\right)^n}{dt\over t}+\int_{\left(1+{x\over n}\right)^n}^{e^x}{dt\over t} \right)=x$$
A: 
Wikipedia: (...) its definition as the unique function
  which is equal to its derivative and is equal to $1$ when $x=0$. That
  is, $$\frac{d}{dx}e^x = e^x \quad\text{and}\quad e^0=1$$

$$\ln{e^x} = \int_{1}^{e^{x}} \frac{1}{s} ds$$
$$ \frac{d}{dx}\ln{e^{x}}=1$$
$$ \ln{e^x} = x + c$$
Since
$$\ln e^0 = \int_{1}^{e^{0}} \frac{1}{s} ds = \int_{1}^{1} \frac{1}{s} ds= 0$$
$c=0$
