Assume that the common distribution $P_{(Z_1,..., Z_n)}$ of $n$ random variables $Z_i: \Omega \to [0 , \infty)$ is distributed as order statistic $(Z_{(1)},..., Z_{(n)})$. That means that for every measurable $A \in \mathcal{B}([0, \infty)^n)$ with respect the product Borel sigma algebra we have
$$ P_{(Z_1,..., Z_n)}(A) = \int_A n! \cdot 1_{(0 \le t_1 \le ... \le t_n \le t)} dt_1 \cdot ... \cdot dt_n $$
with indicator function $1_{(0 \le t_1 \le ... \le t_n \le t)}$ mapping all $n$-tupels $(t_1, t_2,..., t_n)$ to $1$ if $t_1 \le ... \le t_n$, otherwise to zero.
Let $E \in \mathcal{B}[0, \infty)$ any measurable set we define a new random variable $N_n(E): \Omega \to \mathbb{N}$ via
$$ N_n(E)(\omega):= \#\{i \ge 0: Z_i(\omega) \in E \} = \sum_{i=0}^{n} 1_{Z_i \in B}. $$
Question: Why is $N_n(E)$ has binomial distribution $B(n, p)$ where $p= \int_E dt_1$?
That is we have to check that for $1 \le k \le n$
$$ P(N_n(E)=k)= \frac{n!}{k! \cdot (n-k)!}p^k(1-p)^{n-k} $$
In particular I not how to obtain the factor $\frac{1}{k! \cdot (n-k)!}$. My idea was to observe that
$$ P(N_n(E)=k) = \int_{B_k} n! \cdot 1_{(0 \le t_1 \le ... \le t_n \le t)} dt_1 \cdot ... \cdot dt_n $$
where $B_k $ is the union of disjoint measurable sets $ E^{m_1} \times (E^c)^{\overline{m}_1} \times E^{m_2} \times (E^c)^{\overline{m}_2} \times ... E^{m_n} \times (E^c)^{\overline{m}_n} \times $ where the union runs over all $2n$-tupels $(m_1, \overline{m}_1, m_2, \overline{m}_2,...,m_n, \overline{m}_n) $ with $\sum_{i=1}^n m_i =k$ and $\sum_{i=1}^n \overline{m}_i =n-k$.
Why does this gives $\frac{n!}{k! \cdot (n-k)!}p^k(1-p)^{n-k}$?
The question is motivated by a step in the proof of Lemma 2.2 (page 37) proof from Kyprianou's "Introductory Lectures On Fluctuations Levy Processes with Applications" which I not understand.
I'm not very fluent in the 'techincal details' so please let me know if this answer is not convincing enough or misses something important. You can start by considering the sets of the type $E=[0,a)$.
Pick any set $E$ of the type $[0,a)$, then $P\left(N_{n}(E)=k\right)$ is equal to the probability of $\left(Z_{(1)},\ldots,Z_{(k)}\leq a\right)$. Integrating from $dt_{1}$ to $dt_{(k)}$ will yield (please forget about the technical details) $n!\frac{p^{k}}{k!}$: $$ \int_{0}^{t_{2}}n!dt_{1}\left(\int\cdots\right) = \int_{0}^{t_{3}}t_{2}dt_{2}\left(\int\cdots\right) = \int_{0}^{t_{4}}\frac{t_{3}^{2}}{2}dt_{3}\left(\int\cdots\right)=\int_{0}^{t_{5}}\frac{t_{4}^{3}}{3!}\left(\int\cdots\right) =\cdots=\int_{0}^{a}\frac{t_{k}^{k-1}}{(k-1)!}dt_{k} = \frac{p^{k}}{k!} $$
. Now integrating from $dt_{(k+1)}$ up to $dt_{(n)}$ will yield $n!\frac{p^{k}}{k!}\frac{(1-p)^{n-k}}{(n-k)!}$. You can do the same but in this case integrate 'forwards' and end up with the integral between $a$ and $1$ for $\left(1-p\right)^{k-1}/(k-1)!)$ with a flipped sign if necessary.
Inuitively, you have $k$ random variables in $E$ and $n-k$ not in $E$, so you have to count all possible configurations (hence the $\frac{n!}{k!(n-k)!}$).
I think (but am not sure) that without loss of generality you can assume that for $N_{n}(E)=k$ you can assume that it is $\left\{Z_{(1)},\ldots,Z_{(k)}\right\}\in E$, so the proof would follow through.