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 $$q_{n+1}(t)=\frac{k^{n+1}\cdot {t}^{n+1}}{(n+1)\cdot n!}=\frac{(k\cdot t)^{n+1}}{(n+1)!}$$