From the definition of the function \(p_0(t)\) follows \[p_0(t+\Delta t)=p_0(t)\cdot (1-k\cdot \Delta t)\] After all, the probability that the neuron does not fire in the interval \([0,t+\Delta t]\) is equal to the product of the probability that the neuron does not fire in the interval \([0,t]\) and the probability that the neuron does not fire in the interval \([t,t+\Delta t]\).

So we know: \[\frac{p_0(t+\Delta t)-p_0(t)}{\Delta t} =-k\cdot p_0(t)\] When choosing \(\Delta t\) smaller and smaller, we obtain in the limiting case \(\Delta t\rightarrow 0\) resulting in the differential equation \[p_0'(t)=-k\cdot p_0(t)\] At time \(t=0\), the neuron has not yet been able to fire; thus \(p_0(0)=1\).

The function \(p_0(t)\) is therefore a solution of the initial value problem \[p_0'(t)=-k\cdot p_0(t),\quad p_0(0)=1\] but this is part of an exponential decay process, and so the solution is equal to \[p_0(t)=e^{-k\cdot t}\]

From the definition of the function \(p_n(t)\) follows \[p_n(t+\Delta t)=p_n(t)\cdot (1-k\cdot \Delta t)+p_{n-1}(t)\cdot k\cdot \Delta t \] After all, the probability that the neuron fires \(n\) times in the interval \([0,t+\Delta t]\) is equal to the product of the probability that the neuron fires \(n\) times in the interval \([0,t]\) and not in the interval \([t,t+\Delta t]\) plus the probability that the neuron fires \(n-1\) time in the interval \([0,t]\) and once in the interval \([t,t+\Delta t]\).

So we know: \[\frac{p_n(t+\Delta t)-p_n(t)}{\Delta t} =-k\cdot p_n(t)+ k \cdot p_{n-1}(t)\] When we choose \(\Delta t\) smaller and smaller, then we obtain the limiting case \(\Delta t\rightarrow 0\) which results in the differential equation \[p_n'(t)=-k\cdot p_n(t)+ k \cdot p_{n-1}(t)\] At time \(t=0\) the neuron has not yet been able to fire so \(p_n(0)=0\).

The explicit formula for \(p_n(t)\) can be obtained by the method of induction. But is more convenient to define \[q_n(t)=p_n(t)\cdot e^{k\cdot t}\] and verify (do it yourself!) that this results in \[q_n'(t)=k\cdot q_{n-1}(t)\quad\text{with}\quad q_n(0)=0\] and then prove that \[q_n(t)=\frac{(k\cdot t)^n}{n!}\]

For \(n=1\) we have \[q_1'(t)=k\cdot q_0(t)=k\quad\text{with}\quad q_1(0)=0\] so \[q_1(t)=k\cdot t\]

For \(n=2\) we have \[q_2'(t)=k\cdot q_1(t)=k^2t\quad\text{with}\quad q_2(0)=0\] so \[q_2(t)=\frac{(k\cdot t)^2}{2!}\]

Suppose that the assertion is true for \(n\), then this also applies to \(n+1\). After all \[q_{n+1}'(t)=k\cdot q_n(t)=k\cdot \frac{(k\cdot t)^n}{n!} \quad\text{with}\quad q_{n+1}(0)=0\] and thus \[q_{n+1}(t)=\frac{k^{n+1}\cdot {t}^{n+1}}{(n+1)\cdot n!}=\frac{(k\cdot t)^{n+1}}{(n+1)!}\]