exponential distribution I didn't find a demonstration on the web of how to obtain the exponential probability distribution  without using a Poisson distribution, and I don't manage to do it myself.
I am trying to understand how to get this pdf with the only assumption that for each interval dt, the probability of an event is constant. I guess the demo is pretty immediate, but I am missing it.
 A: I still not sure that understood correctly your condition with $dt$, but there is a usual motivation. We want $\tau\in\mathcal E$ not to have memory, i.e.
$$
P(\tau\geq t+s|\tau \geq t) = P(\tau\geq s)
$$
for all $s,t\geq 0$. By the formula of the conditional probability this leads to the fact
$$
P(\tau\geq t+s) = P(\tau\geq s)P(\tau\geq t)
$$
which clearly implies that the density is $\lambda \mathrm e^{-\lambda t}$.
I can make a guess that you also mean the property
$$
\frac{P(\tau \in dt|\tau\geq t)}{dt} = c.
$$
Again, using the formula $\displaystyle{P(A|B) = \frac{P(A\cap B)}{P(B)}}$ we obtain
$$
\frac{P(\tau \in dt)}{dt} = cP(\tau\geq t),
$$
i.e.
$$
f(t) = c(1-F(t))
$$
where $f(t)$ is a density function and $F(t)$ is a distribution function. You only need now to recall that $F' = f$ and solve an ODE.
A: The question is not clear, but maybe you'll find the following useful.
For $t > 0$ fixed, let $E_k$, for $k=1,\ldots,n$, denote the event of $0$ events in the time interval 
$I_k = [\frac{{(k - 1)t}}{n},\frac{{kt}}{n}]$. Since $I_k$ has length $t/n$,
$$
{\rm P}(E_k^c) = \lambda \frac{{ t}}{n} + o\bigg(\frac{1}{n}\bigg) \;\; {\rm as} \;\; n \to \infty.
$$
Hence
$$
{\rm P}(E_k) = 1 - \lambda \frac{{ t}}{n} + o\bigg(\frac{1}{n}\bigg) \;\; {\rm as} \;\; n \to \infty.
$$
Now, by independence, the probability $p_t$ of $0$ events in the time interval $[0,t]$ is given by
$$
p_t = {\rm P}(E_1 ) \cdots {\rm P}(E_n ) = \bigg(1 - \frac{{\lambda t}}{n} + o\bigg(\frac{1}{n}\bigg)\bigg)^n .
$$
Letting $n \to \infty$, this gives
$$
p_t = e^{ - \lambda t} .
$$
This probability corresponds to ${\rm P}(X > t)$, where $X$ is exponential with parameter $\lambda$.
EDIT: Note that $n o(1/n) \to 0$ as $n \to \infty$. Using this, it indeed follows that 
$$
\bigg(1 - \frac{{\lambda t}}{n} + o\bigg(\frac{1}{n}\bigg)\bigg)^n  \to e^{ - \lambda t}  .
$$
