Homogeneous Poisson Process of spike train In Theoretical Neuroscience by Dayan and Abbott page 25, 


  the probability $P[t_1, t_2, ..., t_n]$ that a sequence of n spikes occurs with spike $i$ falling between times $t_i$ and $t_i+\Delta t$ for $i=1,2,...,n$ is given in terms of this by the relation $P[t_1, t_2, ..., t_n]=p[t_1,...,t_n](\Delta t)^n$... the probability $P[t_1, t_2, ..., t_n]$ can be expressed in terms of another probability function $P_T[n]$, which is the probability that an arbirary sequence of exactly n spikes occurs within a trial of duration T. Assuming that the spike times are ordered so that $0 \le t_1 \le t_2 \le ... \le t_n \le T$, the relationship is 
  $$P[t_1, t_2, ..., t_n] = n!P_T[n]\big({\Delta t \over T}\big)^n$$


I get that $P_T[n]$ is the probability function of $n$ spikes occurring(which later is proved to be a Poisson distribution).  I'm guessing $n!$ is different ways of re-arranging the $n$ spikes. What exactly is $\big({\Delta t \over T}\big)^n$?
First of all you should notice the assumption that spikes are independent and that every sequence of spikes over an interval has the same probability. That's why you can express $P[t_{1}, t_{2}, ..., t_{n}]$ in terms of $P_{T}[n]$. 
Second, note that $p[t_{1}, t_{2}, ..., t_{n}]$ is a probability density. So you can express the probability as the integral of $p[t_{1}, t_{2}, ..., t_{n}] dt_{1}dt_{2}...dt_{n}$

$\Delta t / T$ is the probability that one spike occurs in that time interval. If $\Delta t$ is small enough, only one spike is within that interval, so just let $\Delta t \rightarrow dt_{i}$. Then  $p[t_{1}, t_{2}, ..., t_{n}] dt_{1}dt{2}...dt{n} = n!P_{T}[n]dt_{1}dt_{2}...dt_{n}(\frac{1}{T})^{n}$

You should see this as a joint probability.