Here I record two ways to derive Poisson distribution. A Poisson process is the number of events $\{N(t),t\geq 0\}$ in some given time $[0,t]$, with the following property:

• A point process with $0$ origin

$$\begin{align} N(0)=0,\quad N(t)\geq 0,\, \forall t \end{align}$$

• Independent increment

$$\begin{align} &N(t_1)\perp\!\!\!\perp N(t_2)-N(t_1)\perp\!\!\!\perp,\ldots,\perp\!\!\!\perp N(t_m)-N(t_{m-1}),\\ &\quad \forall\, 0<t_1<t_2<\ldots<t_m ,\quad \forall m \end{align}$$

• Stationary increment

$$\begin{align} \mathbb{P}\left( N(s+t)-N(s)=n \right) =\mathbb{P}\left( N(t)=n \right) ,\quad \forall\, s,t\geq 0,\,n\geq 0 \end{align}$$

• Linear increment: for $\delta t$ small enough

$$\begin{align} \begin{cases} \mathbb{P}\left( N(t+\delta t)-N(t)=1 \right)=\lambda \delta t+o(\delta t)\\ \mathbb{P}\left( N(t+\delta t)-N(t)\geq 2 \right) =o(\delta t) \end{cases} \end{align}$$

Distribution of $N(t)$:

$$\begin{align} \mathbb{P}\left( N(s+t)-N(s)=k \right)=\mathbb{P}\left( N(t)=k \right)=\dfrac{(\lambda t)^k}{k!} e^{-\lambda t}\sim P(\lambda t) \end{align}$$

Proof 1

We divide $[0,t]$ into $M$ pieces with $M\to \infty$:

$$\begin{align} \mathbb{P}\left( N(t)=k \right)=&\lim_{M\to \infty}\binom{M}{k}\left( \lambda \dfrac{t}{M} \right)^k\left(1-\lambda \dfrac{t}{M}\right)^{M-k} \\ =&\dfrac{(\lambda t)^k}{k!}\lim_{M\to \infty}\left[\left( 1-\dfrac{\lambda t}{M} \right)^{M/\lambda t}\right]^{\lambda t}\dfrac{M!}{(M-k)!(M-\lambda t)^k}\\ =&\dfrac{(\lambda t)^k}{k!}e^{-\lambda t} \end{align}$$

Note: this proof is not quite rigorous, but it's straight enough. I would say it's rather a "physicist's method" (lol).

Proof 2

For convenience, denote that $\mathbb{P}\left( N(t)=n \right):=P_n(t)$. For a small $h\to 0$, we consider the process from $t\to t+h$. First focus on a special case $n=0$:

$$\begin{aligned} P_0(t+h)=&\mathbb{P}\left( N(t+h)=0 \right)\\ =&\mathbb{P}\left( N(t)=0 \right) (1-\lambda h) \end{aligned}$$

i.e. note that $P_0(0)=1$

$$\begin{align} P_0'(t)=\lim_{h\to 0}\dfrac{P_0(t+h)-P_0(t)}{h}=-\lambda \Rightarrow P_0(t)=e^{-\lambda t} \end{align}$$

Then focus on $n\geq 1$. We could use mathematical induction to obtain all $n\geq 1$ cases:

$$\begin{align} \{N(t+h)=n\}=&\{N(t)=n,\,N(t+h)-N(t)=0\}\\ &\bigcup \{N(t)=n-1,\,N(t+h)-N(t)=1\}\\ &\bigcup_{l=2}^n \{N(t)=n-l,\,N(t+h)-N(t)=l\} \end{align}$$

those $l\geq 2$ term would be with $\mathbb{P}(\, \cdot \, )\sim o(h)$. Then

$$\begin{align} \mathbb{P}\left( N(t+h)=n \right)=& \mathbb{P}\left( N(t)=n \right)(1-\lambda h)+\mathbb{P}\left( N(t)=n-1 \right)\lambda h \end{align}$$

i.e.

$$\begin{align} P_n'(t)=\lim_{h\to 0}\dfrac{\mathbb{P}\left( N(t+h)=n \right)-\mathbb{P}\left( N(t)=n \right) }{h}=-\lambda P_n(t)+\lambda P_{n-1}(t) \end{align}$$

with initial condition

$$\begin{align} P_n(0)=\mathbb{P}\left( N(0)=n \right) =\delta _{n,0} \end{align}$$

Solution to the above differential equation is

$$\begin{align} &\dfrac{\mathrm{d}^{} e^{\lambda t}P_n(t)}{\mathrm{d}t^{}}=\lambda e^{\lambda t}P_{n-1}(t),\quad P_0(t)=e^{-\lambda t}\\ \Rightarrow &\mathbb{P}\left( N(t)=k \right)= P_k(t)=\dfrac{(\lambda t)^{k}}{k!}e^{-\lambda t} \end{align}$$