I'm trying to prove that for a Poisson process $N_t$ ,$ \ N_t \sim \text{Poisson}(\lambda t) $ for $t\geq 0$. I've got to the point where I have found the differential equation $p_{n}^\prime(t) = -\lambda p_n(t) + \lambda p_{n-1}(t) $ where $p_n(t) = P(N_t = n) $.
How can I solve this equation and show that $p_n(t) = \frac{e^{-\lambda t} (\lambda t)^n}{n!} $ and thus that $N_t$ is Poisson distributed?
Here's my working so far:
$$ \begin{align*} p_n(t+h) & = P(N_{t+h} = n) \\ & = P(N_{t+h} = n| N_t > n) P(N_t > n) + P(N_{t+h} = n | N_t = n)P(N_t = n)\\ &+ P(N_{t+h} = n | N_t = n-1) P(N_t = n-1) + P(N_{t+h} = n |N_t < n-1)P(N_t < n-1) \\ &= (1-\lambda h + o(h))p_{n}(t) + (\lambda h + o(h))p_{n-1}(t) + o(h) \end{align*} \implies \frac{p_{n}(t+h) - p_n(t)}{h} = -\lambda p_n(t) + \lambda p_{n-1}(t) + \frac{o(h)}{h}$$ Taking $h \rightarrow 0 $ we get $p_{n}^\prime(t) = -\lambda p_n(t) + \lambda p_{n-1}(t) $.