In the book: K. Itô, Essentials of Stochastic Processes, American Mathematical Society(Translation of Mathematical Monographs; v.231), Providence, RI.(2006) Th 2.9.2-3, p.33-35, there are following theorems:
Theorem 2.9.2. If $X_t, t\in T=[a,b) $ is a separable Poisson process in wide sense, then with probability 1, its sample process is a step function which increases only by jumps of magnitude 1.
Theorem 2.9.3. If an additive process which is continuous in probability has the property that, with probability 1, its sample process is a step function
which increasing only with jumps of magnitude 1, then it is a separable Poisson process in a wide sense.
In Ito's book, the additive process $X_t$-- process with independent increments and $X_0=0$.
Poisson process in a wide sense -- a process $X_t$ with independent increments, which is continuous in probability and the distribution of $X_t-X_s$ for each pair $t>s$ is a Poisson distribution.
From above theorems, if $N=\{N_t, t\ge 0\}$ is a stochastic continuous process, then
the $N$ is Poisson process in a wide sense(i.e., nonhomogeneous Poisson process) may be
characteristicrized as $N$ is a simple counting process with independent increments,
where simple counting process means that its sample process is a step function
which increasing only with jumps of magnitude 1.
Now, if $N$ is a nonhomogeneous Poisson process, then $N_t-N_s$ is a Poisson distributed r.v. for each pair $t>s$ and
\begin{equation*}
m(t)=\mathsf{E}[N_t] \tag{1}
\end{equation*}
is a non-negative increasing continuous function of $t$. Furthermore, the $N_t-N_s$ is
Poisson distributed as
\begin{equation*}
\mathsf{P}(N_{t+s}-N_t=k)=\frac{[m(t+s)-m(t)]^k}{k!}e^{-[m(t+s)-m(t)]}, \qquad k=0,1,2,\cdots.
\end{equation*}
If $m(t)$ in (1) is an absolutely continuous function, or epuivalently, the L-S measure $\mathrm{d}m$ generated by $m$ is absolute continuous with respect to the Lebesgue measure $\mathrm{d}t$,
and $\lambda(t)=\frac{dm}{dt}(t)$, then
\begin{equation}
m(t)=\int_{0}^{t}\lambda(s)\,\mathrm{d}s.
\end{equation}
Best Answer
Variables $s$ and $t$ are the other way around in Ross's book so I'll go with that. So we're trying to prove:
$$P(N(s+t)-N(s)=n)=e^{-(m(s+t)-m(s))}\dfrac{[m(s+t)-m(s)]^n}{n!},\quad n\geq 0\qquad\qquad(*)$$
For easier notation, let $P_n(u,v) := P(N(v)-N(u)=n),\;$ for $n\geq 0\;$ and $u\leq v$.
Firstly, we take the case $n=0$.
\begin{eqnarray*} P_0(s,s+t+h) &=& P(N(s+t+h)-N(s)=0) \\ &=& P(N(s+t)-N(s)=0)\; P(N(s+t+h)-N(s+t)=0) \\ &=& P_0(s,s+t)\; P_0(s+t,s+t+h) \\ &=& P_0(s,s+t)\; (1-\lambda(s+t)h+o(h)) \\ P_0(s,s+t+h) - P_0(s,s+t) &=& -\lambda(s+t)hP_0(s,s+t)+o(h) \\ P_0^{'}(s,s+t) &=& -\lambda(s+t)P_0(s,s+t) \qquad\quad\text{(divide by $h$ and let $h\to 0$)} \\ \therefore\quad P_0(s,s+t) &=& Ce^{-m(s+t)} \qquad\text{since $\frac{d}{dt}e^{-m(s+t)} = -\lambda (s+t)e^{-m(s+t)}$}. \\ && \\ P_0(s,s)=1\quad \text{ so }\quad 1 &=& Ce^{-m(s)} \\ C &=& e^{m(s)} \\ && \\ \therefore\quad P_0(s,s+t) &=& e^{-(m(s+t)-m(s))}. \end{eqnarray*}
Now we take the general case, $n\gt 0$.
\begin{eqnarray*} P_n(s,s+t+h) &=& P(N(s+t+h)-N(s)=n) \\ &=& \sum_{i=0}^{n} P(N(s+t)-N(s)=i)\; P(N(s+t+h)-N(s+t)=n-i) \\ &=& \sum_{i=0}^{n} P_i(s,s+t)\; P_{n-i}(s+t,s+t+h) \\ &=& P_n(s,s+t)\; (1-\lambda(s+t)h+o(h)) + P_{n-1}(s,s+t)\; (\lambda(s+t)h+o(h)) + o(h) \\ \end{eqnarray*} \begin{eqnarray*} P_n(s,s+t+h) - P_n(s,s+t) &=& -\lambda(s+t)hP_n(s,s+t) + \lambda(s+t)hP_{n-1}(s,s+t) + o(h) \\ P_n^{'}(s,s+t) &=& \lambda(s+t)(P_{n-1}(s,s+t) - P_n(s,s+t)) \\ \end{eqnarray*} \begin{eqnarray*} \therefore\quad e^{m(s+t)}\left[ P_n^{'}(s,s+t) + \lambda (s+t)P_n(s,s+t) \right] &=& \lambda (s+t) e^{m(s+t)} P_{n-1}(s,s+t) \\ \frac{d}{dt}\left[ e^{m(s+t)} P_n(s,s+t) \right] &=& \lambda (s+t) e^{m(s+t)} P_{n-1}(s,s+t). \end{eqnarray*}
From here we can prove the result by induction on $n$. The initial case for $P_0(s,s+t)$ has already been proved. So we assume that $(*)$ holds for $n-1$.
\begin{eqnarray*} \frac{d}{dt}\left[ e^{m(s+t)} P_n(s,s+t) \right] &=& \lambda (s+t) e^{m(s+t)} e^{-(m(s+t)-m(s))}\dfrac{[m(s+t)-m(s)]^{n-1}}{(n-1)!} \\ &=& \lambda (s+t) e^{m(s)}\dfrac{[m(s+t)-m(s)]^{n-1}}{(n-1)!} \\ e^{m(s+t)} P_n(s,s+t) &=& C + e^{m(s)} \dfrac{[m(s+t)-m(s)]^{n}}{n!}. \end{eqnarray*}
Setting $t=0$ to evaluate the constant $C$ we get $C=0$ and the proof is complete as we arrive at
$$P_n(s,s+t) = e^{-(m(s+t)-m(s))} \dfrac{[m(s+t)-m(s)]^{n}}{n!}.$$