Find a complete sufficient statistics, or show that one does not exist $$  f(x|\theta) = e^{x-\theta}\exp\left(-e^{x-\theta}\right),\;\;\; -\infty < x < \infty,\;\;\; -\infty < \theta < \infty.$$
Find  acompliete sufficient statistics, or show that one does not exists. 

What I have found was that this given $f$ is not an exponential family, and my solution says as followed; 
There is no complete sufficient statistics for that. In detail, the solution said the order statistics are minimal sufficient, and this is location family. Thus, the range $R = X_{(n)} - X_{(1)}$ is ancilliary, and expectation does not depend on $\theta$. So this sufficient statistics is not complete. 
I don't understand two points. First, the solution could be the proof for no-existency of complete statistics, because this is only the proof for the case of $R = X_{(n)} - X_{(1)}$, 
and second, The fact that the expectation does not depend on $\theta$ could imply that this sufficient statistics is not complete. 
Could anybody help me to understand this? 
 A: Contrary to your solution that says "There is no complete sufficient statistic...", I claim the following:

Claim: Given a random sample $\mathbf{X} = (X_1,\ldots,X_n)$ from the pdf
  $$f(x|\theta) = e^{x-\theta}\exp\left(-e^{x-\theta}\right),\;\;\; -\infty < x < \infty,\;\;\; -\infty < \theta < \infty\,, $$ 
  the statistic defined by $T(\mathbf{X}) = \sum\limits_{i=1}^{n} \exp{(X_i)}$ is a complete sufficient statistic for $\theta$.

Proof:
Sufficiency of $T(\mathbf{X})$. Rewriting the pdf,
$$ 
f(x|\theta) = e^{-\theta}\left[\exp(-e^{-\theta})\right]^{e^{x}}\cdot e^{x}
$$
resulting to the likelihood
$$
f(\mathbf{x}|\theta) = e^{-n\theta}\left[\exp(-e^{-\theta})\right]^{T(\mathbf{x})}\cdot \exp\left(\sum\limits_{i=1}^{n}x_i\right)\,.
$$
Thus, by the Factorization Theorem, $T(\mathbf{X})$ is sufficient.
Completeness of $T(\mathbf{X})$. Showing that $T(\mathbf{X})$ is a complete statistic is equivalent to showing that the family of its pdf is complete. By simple transformation, it easy to see that $\exp(X_i)\sim \text{exponential}(\beta)$ with $\beta = e^{\theta}$ for every $i=1,\ldots,n$, implying 
$$T(\mathbf{X}) = \sum_{i=1}^{n} \exp{(X_i)} \sim \text{Gamma}(n,e^{\theta})$$
where we note that $\exp(X_i)$'s are also i.i.d.
The conclusion follows immediately from the fact that the family of gamma distributions is complete as a result of the uniqueness of Laplace transforms.
