Q:

Can I use the maximum likelihood to show a statistic is complete?

Problem: Consider a random sample of size $n$ that follows a  density probability function given by:

$$f(x,\theta)=\frac{1}{\theta} x^{-\frac{\theta+1}{\theta}}\mathbb{1}_{(1,+\infty)},\:\:\theta>0$$

where $\theta$ is unknown.
Provide an sufficient statistic for this model.

In another thread I was able with the help of mathstckaexchenge members to derive the following         maximum likelihood estimator $\hat{\theta} = \left( \frac{1}{n} \sum\limits_{i=1}^n \ln x_i \right) + 1$
I want to use the statistic obtained to answer the question of sufficiency.
For that I am going to use the factorization theorem, which states that a statistic is sufficient for $\theta$ iff 
$$f(x|\theta)=g(T(x)|\theta)h(x)$$
However to compute the joint distribution I would use the same procedure as to compute the 
likelihood function and I would obtain:
$f(x|\theta)=\prod_\limits{i=1}^{n}f(x_i,\theta)=\frac{1}{{\theta}^n}\prod_\limits{i=1}^{n} x_i^{-\frac{{(\theta+1)}}{{\theta}}}\mathbb{1}_{(1,+\infty)}$ 
However I am not able to separate the the x_{is}
 from $\theta$ if I do not use the logarithm.
Questions:
How do I solve the problem? Is it possible to use the maximum likelihood function for sufficiency? Can I use the logarithm? If so why?
Thanks in advance!

A:

I hope I understood your question correctly after the clarification in the comments.
Indeed you can use logarithms if you want to, and you don’t need to consider $\log f$ instead of $f$ in order to do so. Since
$$
f(x\mid\theta)=\prod_{i=1}^nf(x_i,\theta)=\frac1{\theta^n}\prod_{i=1}^n x_i^{-\frac{\theta+1}\theta}=\frac1{\theta^n}\exp\left(-\frac{\theta+1}\theta\sum_{i=1}^n\ln x_i\right)\;,
$$
we have
$$
f(x\mid\theta)=g(T(x)\mid\theta)h(x)\;,
$$
with
$$
T(x)=\sum_{i=1}^n\ln x\;.
$$
(Note that I didn’t include the indicator function, which isn’t needed here (nor in your maximum likelihood estimator) because the samples lie in $[1,\infty$] anyway.)
Thus $T(x)$ is a sufficient statistic.
But you don’t have to use logarithms. You could also just write
$$
f(x\mid\theta)=\prod_{i=1}^nf(x_i,\theta)=\frac1{\theta^n}\prod_{i=1}^n x_i^{-\frac{\theta+1}\theta}=\frac1{\theta^n}\left(\prod_{i=1}^n x_i\right)^{-\frac{\theta+1}\theta}\;,
$$
which factorizes with $S(x)=\prod_{i=1}^nx_i$. That $S(x)$ and $T(x)$ are likewise sufficient statistics is not surprising, since one is an injective function of the other ($S(x)=\exp(T(x))$ and $T(x)=\log(S(x))$), and an injective function of a sufficient statistic is again a sufficient statistic.

