We know that a homogeneous Poisson process is a process with a constant intensity $\lambda$. That is, for any time interval $[t, t+\Delta t]$, $P\left \{ k \;\text{events in}\; [t, t+\Delta t] \right \}=\frac{\text{exp}(-\lambda \Delta t)(\lambda \Delta t)^k}{k!}$.
And therefore, event count in $[0, T]$ follows a Poisson distribution with rate $\lambda T$. That is, $P\left \{ N(T)=k\right \}=\frac{\text{exp}(-\lambda T)(\lambda T)^k}{k!}$. ($N$ is the count.)
The problem is:
Prove that the following simulation generates a homogeneous Poisson process with rate $\lambda$ on $[0, T]$: Step 1: Sample $k$ from Poisson distribution with mean $\lambda T$. Step 2: Sample $s_1, \cdots,s_k$ i.i.d. from uniform $[0, T]$. That is, demonstrate that for any time interval $[t, t+\Delta t]$ in $[0,T]$, $P\left \{ k \;\text{events in}\; [t, t+\Delta t] \right \}=\frac{\text{exp}(-\lambda \Delta t)(\lambda \Delta t)^k}{k!}$.
My question is is this problem trivial? Because we have that $k$ is sampled from Poisson distribution with mean $\lambda T$, so $k$ follows a Poisson distribution with rate $\lambda T$. So $k$ is the count in $[0,T]$. And for any time interval $[t, t+\Delta t]$ in $[0,T]$, we can treat any interval as $[0,T']$ with a new length $T'$, so we replace $[0,T]$ with $[t, t+\Delta t]$ and $T$ becomes $\Delta T$, so we then have $P\left \{ k \;\text{events in}\; [t, t+\Delta t] \right \}=\frac{\text{exp}(-\lambda \Delta t)(\lambda \Delta t)^k}{k!}$. Am I right? Is it trivial or my proof is weak?
Thanks in advance.