$\prod_{i=1}^n[(N-x_i)!]=\prod_{i=1}^n[(N-y_i)!]$ for all $N\in\mathbb{Z}^+$ implies $x_{(i)}=y_{(i)}$, for all $i\in\{1,\cdots,n\}$ Suppose $$\prod_{i=1}^n[(N-x_i)!]=\prod_{i=1}^n[(N-y_i)!]$$ for all $N\in\mathbb{Z}^+$.
Why does it imply $x_{(i)}=y_{(i)}$, for all $i\in\{1,\cdots,n\}$? Here $X_{(i)}$ is the order statistic.
I am trying to find a minimal sufficient statistic for $\{X_i\}_{i=1}^n\sim\text{Binomial}(N,\frac{1}{2})$. The last step is showing $\prod_{i=1}^n[(N-x_i)!]=\prod_{i=1}^n[(N-y_i)!] \implies x_{(i)}=y_{(i)}$.
 A: \begin{equation} \prod_{i=1}^n[(N-x_i)!]=\prod_{i=1}^n[(N-y_i)!]\iff\prod_{i=1}^n[(N-x_{(i)})!]=\prod_{i=1}^n[(N-y_{(i)})!] \end{equation} $(N-x_{(i)})!$ is a polynomial of $N$ to the degree of $N-x_{(i)}$.
$\prod_{i=1}^n[(N-x_{(i)})!]$ is a polynomial of $N$ to the degree of $nN-\sum_{i=1}^nx_{(i)}$. Its powers include $nN-\sum_{i=1}^{n} x_{(i)}$, $(n-1)N-\sum_{i=1}^{n-1} x_{(i)}$, $(n-2)N-\sum_{i=1}^{n-2} x_{(i)},\cdots$.
This means $$\sum_{i=1}^m x_{(i)}=\sum_{i=1}^m y_{(i)}, \forall m\in\{1,\cdots,n\}\iff x_{(i)}=y_{(i)},\forall i\in\{1,\cdots,n\}$$
--- Update:
$(N-x_{(i)})!$ is like $N^{N-x_{(i)}}+c_{i,1}N^{N-x_{(i)}-1}+c_{i,2}N^{N-x_{(i)}-2}+\cdots$
$\prod_{i=1}^n[(N-x_{(i)})!]$ is like $(N^{N-x_{(n)}}+c_{n,1}N^{N-x_{(n)}-1}+c_{n,2}N^{N-x_{(n)}-2}+\cdots)\times(N^{N-x_{(n-1)}}+c_{n-1,1}N^{N-x_{(n-1)}-1}+c_{n-1,2}N^{N-x_{(n-1)}-2}+\cdots)\times(N^{N-x_{(n-2)}}+c_{n-2,1}N^{N-x_{(n-2)}-1}+c_{n-2,2}N^{N-x_{(n-2)}-2}+\cdots)\underset{\text{n terms product}}{\cdots}$
Thus the largest degree term of $\prod_{i=1}^n[(N-x_{(i)})!]$ is $N^{nN+\sum_{i=1}^nx_{(i)}}$. Also it has terms like  $N^{(n-1)N+\sum_{i=1}^{n-1}x_{(i)}}$, $N^{(n-2)N+\sum_{i=1}^{n-2}x_{(i)}}$, $\cdots$
If $\prod_{i=1}^n[(N-x_{(i)})!]=\prod_{i=1}^n[(N-y_{(i)})!],\forall N\in\{1,2,\dots\}$, their terms must match. Thus this implies $$\sum_{i=1}^mx_{(i)}=\sum_{i=1}^my_{(i)},\forall m\in\{1,2,\dots\,n\}$$
