Why is the first postulate of the Poisson process that $\lambda dt$ is the probability of exactly one event in $[t,t+dt]$? From the pdf of the Poisson distribution I would expect $\Pr(x=1)$ to be 
$$\lambda dt \cdot \exp(-\lambda dt)$$
I can see that as $dt$ gets very small, $\exp(-\lambda dt)$ becomes close to $1$, and so suggests $\lambda dt$, but I don't see why in the limit as $dt \to 0$ of $\lambda dt \cdot \exp(-\lambda dt)$, doesn't that make the $\lambda dt$ term zero also?
 A: In fact, Leibniz' notation for infinitesimal increments can be confusing.
One has to be careful here to keep all terms of the same order: $e^{-\lambda dt}$ must be approximated to first order of $dt$ (not zeroth order, i.e. without any terms in $dt$), i.e. $e^{-\lambda dt}$ is approximately $1 - \lambda  dt$ + (plus terms which are at least quadratic in $dt$ and thus go to zero faster than $dt$ itself)
Then one has:
$$
    \lambda dt \cdot e^{-\lambda dt} \approx \lambda dt \cdot (1 - \lambda dt)
$$
and then for $dt \rightarrow 0$ one can ignore the non-leading terms ($dt^2$) and is left with $\lambda dt$. 
A: The previous two answer's are I think coming at the problem "backwards" - though they are both correct.  They do not start with the postulate and end with the conclusion.  If we start from the postulate, then we have:
$$Pr(\text{No event in} [t,t+dt])=1-Pr(\text{1 event in} [t,t+dt])=1-\lambda dt$$
If we define the function $h(t)$ as follows:
$$Pr(\text{No event in} [0,t])=h(t)$$
$$Pr(\text{No event in} [0,t+dt])=h(t+dt)$$
Additionally, we can use the independence of the increments - another postulate of the poisson process and we have:
$$h(t+dt)=h(t)[1-\lambda dt]\implies\frac{h(t+dt)-h(t)}{dt}=-\lambda h(t)$$
Taking the limit as $dt\to 0$ we have $h'(t)=-\lambda h(t)$ which implies $h(t)=K\exp(-\lambda t)$.  We can resolve the proportionality constant by noting that $h(0)=1$ - i.e. it is certain to see no events in $[0,0]$.  This gives $K=1$.  This derivation can be found here (page 4) along with how to extend it to the probability for any number of events (basically by multiplying the zero count probability by $\lambda^n$ where $n$ is the number of events).
A: Here's an alternative (but basically equivalent) derivation to @Andre Holzner's:
For a Poisson process $N(t)$ with rate $\lambda$, 
$Pr(N(t+\tau) - N(t) = 1) = (\tau\lambda)\exp(-\tau\lambda) = Pr(N(\tau) = 1) $
which has Taylor expansion around $\tau=0$
$\tau\lambda - \tau^2\lambda^2 + O(\tau^3)$
and this is approximately $\tau\lambda$ for small $\tau$. You're correct that the actual limit is zero, as one typically assumes $Pr(N(0)=0)=1$ in developing the Poisson process.
