Proof that n-order statistics are sufficient for a sample of size n This is problem 1.5.8 in Mathematical Statistics by Bickel and Doksum. It seems straightforward, but I am not sure if my proof is lacking in some way. It doesn't seem quite correct.

Question Let $X_1, X_2,..., X_n$ be a sample from some continuous distribution $F$ with density $f$, which is unknown. Treating $f$ as a
  parameter, show that the order statistics $X_{(1)},...X_{(n)}$ (cf.
  Problem B.2.8) are sufficient for $f$.

My Solution Attempt
The sample density is simple the product of the $f(x_i)$.
This yields
$$\prod_{i=1}^nf(X_{i})$$
We apply the factorization theorem where $h(x)=1$
$$g(\theta)=\prod_{i=1}^nf(X_{i})$$ 
(here $\theta=X_{(1)},...X_{(n)}$). And so because we have factorized the sample density, the statistics are sufficient.
 A: The joint distribution off all order statistics $X_{(1)}, \dots, X_{(n)}$ is
$$f_o(y_1, \dots, y_n) = n! f(y_1) \times \cdots \times f(y_n)$$
for $y_1 \le \cdots \le y_n$.
Thus, the joint distribution of $X_1 ,\dots, X_n$, given $X_{(1)}, \dots, X_{(n)}$ does not depend of the density $f$! We have
$$ Pr(X_1 = x_1, \dots, X_n= x_n | X_{(1)} = y_1, \dots, X_{(n)} = y_n ) = {1\over n!} $$
when the multisets of the $x_i$'s and of the $y_i$'s are equal.
The intuition is that loosing the drawing order doesn’t matter, for the $X_i$'s are independent.
PS Re-reading this answer long after, I tend to think that the conclusion is clear in itself: given $y_1 < \dots < y_n$,  $Pr(X_1 = x_1, \dots, X_n= x_n | X_{(1)} = y_1, \dots, X_{(n)} = y_n ) = {1\over n!}$ means that all re-orderings of the $y_i$'s are equally probable, whatever the density $f$ is. In fact I don't see how to prove the starting result I used (the joint distribution of order statistics) without proving this at the same time... This just comes from the fact that all the points $x = (x_1, \dots, x_n)$ obtained from a permutation of the $y_i$ have the same density $\prod_i f(x_i) = \prod_i f(y_i)$.
