# Sufficient Statistic for exp{exp{x-theta}}- type density

I'm wondering if this density has a sufficient statistic to summarize an independent and identically distributed sample of random variables $X_1,X_2,...,X_n$ of this density.

The density is:

$f(x|\theta)=e^{-(x-\theta)}e^{-e^{-(x-\theta)}}I_{\mathbb{R}}(x)$

After some computation, arrived to:

$f(\textbf{x}|\theta)=e^{n\theta-\sum_{i=1}^{n} x_i}e^{\sum_{i=1}^{n}(-e^{-(x_i-\theta)})}I_{\mathbb{R}^n}(\textbf{x})$

But with this expression, it seems impossible to resume the information in a statistic because of the exponent of the middle factor. There, the $x_i$'s will allways be separated, but together (each one) with the parameter ($\theta$). So it will be impossible to factorize this expression (to use the Halmos Factorization Theorem), meaning that there is no sufficient statistic. Am I concluding correctly, or i'm missing something?

$$f(\textbf{x}|\theta)=e^{n\theta-\sum_{i=1}^{n} x_i}e^{\sum_{i=1}^{n}(-e^{-(x_i-\theta)})}I_{\mathbb{R}^n}(\textbf{x}) = e^{-\sum_{i=1}^{n} x_i}\cdot e^{n\theta-e^\theta\sum_{i=1}^{n}e^{-x_i}}=h(x)g\left(\theta,\sum_{i=1}^{n}e^{-x_i}\right)$$ By Neyman-Fisher factorization theorem, $T=\sum_{i=1}^{n}e^{-x_i}$ is a sufficient statistic for $\theta$.