Proof. For \(n = 0\): there are no events in \((0, t + h]\) exactly when there are none in \((0, t]\) and none in \((t, t + h]\). By independent and stationary increments, \[P_0(t + h) = P_0(t)\left [1 - \lambda h + o(h)\right ],\qquad \text {so}\qquad \frac {P_0(t + h) - P_0(t)}{h} = -\lambda P_0(t) + \frac {o(h)}{h} .\] Letting \(h\to 0\), \(P_0'(t) = -\lambda P_0(t)\), and with \(P_0(0) = 1\) this gives \(P_0(t) = e^{-\lambda t}\).

For \(n\geq 1\): there are \(n\) events in \((0, t + h]\) if there are \(n\) in \((0, t]\) and none in \((t, t + h]\), or \(n - 1\) and then one, or fewer and then at least two, the last having probability \(o(h)\). So \[P_n(t + h) = P_n(t)(1 - \lambda h) + P_{n-1}(t)\lambda h + o(h),\qquad P_n'(t) = -\lambda P_n(t) + \lambda P_{n-1}(t) .\] Multiplying by \(e^{\lambda t}\), the equation becomes \(\frac {d}{dt}\left [e^{\lambda t}P_n(t)\right ] = \lambda e^{\lambda t}P_{n-1}(t)\). Suppose \(P_{n-1}(t) = \frac {e^{-\lambda t}(\lambda t)^{n-1}}{(n - 1)!}\). Then \(\frac {d}{dt}\left [e^{\lambda t}P_n(t)\right ] = \frac {\lambda ^nt^{n-1}}{(n - 1)!}\), and integrating with \(P_n(0) = 0\) gives \(e^{\lambda t}P_n(t) = \frac {(\lambda t)^n}{n!}\). The result follows by induction on \(n\). The statement for \(N(s + t) - N(s)\) follows from stationarity. □