The problem is to show that the largest order statistic $x_{(n)}$ is sufficient for $\theta$ where $X \sim \frac{1}{\theta}\mathbb{I}_{x \in (0, \theta)}$ is a uniform distribution.
I believe I have proved the result, but my argument cannot make sense, due to the consequences. Here's my argument:
The likelihood of a random sample of size $n$ is, for $\theta >0$, $$L(\theta) = \bigg(\frac{1}{\theta} \bigg)\bigg(\frac{1}{\theta} \bigg) \cdots \bigg(\frac{1}{\theta} \bigg) = \frac{1}{\theta^n},$$ which factors into a product of a function of $\theta$ and $\hat\theta$, and a function of the sample $x_i$ for any statistic $\hat\theta$. Therefore, by the Fisher-Neyman Factorization Theorem, any statistic is sufficient for $\theta$. In particular, the largest order statistic $x_{(n)}$ is sufficient for $\theta$.
It seems to me like any statistic should be sufficient for $\theta$. Of course, "any" statistic includes a constant, so my argument must be wrong. I'm not sure why, though. Can anyone enlighten me?
Thanks