I was looking for a sufficient statistic for a random sample of this distribution when a question came to mind.
$$ f_X(x)=\frac{\left(\theta-1\right)^{x-1}}{\theta^x}, \ \ \ x=1,2,...\ and \ \ \theta >1 $$
This is my approach.
$$ L(\theta) = \prod_{i=1}^{n}\frac{\left(\theta -1\right)^{x_i-1}}{\theta^{x_i}} = \frac{(\theta -1)\strut^{\sum_{i=1}^{n} (x_i)-n}} {\theta\strut^{\sum_{i=1}^{n} x_i}} = h\left(\textstyle{\sum_{i=1}^{n}} x_i;\theta \right)k(x_1,x_2,...,x_n) $$
where $k(x_1,x_2,...,x_n)=1$.
I think that this means $\sum_{i=1}^{n} x_i$ is sufficient for $\theta$. However, I understand that the function $h$ can't have any of the data in it (any $x_i$'s) except in the form of the statistic, which in this case is $\sum_{i=1}^{n} x_i$. But is $h$ allowed to have the sample size $n$ in it like in this example? The sample size could be viewed as a function of the data (e.g. $len(x_1, x_2,..., x_n) =n$), so maybe $n$ shouldn't be in $h$. However, sufficiency has to do with the amount of information that is provided about the parameter $\theta$ so I'm thinking that since $n$ provides no information about $\theta$, $n$ can be in either factor, $h$ or $k$, when doing Fisher–Neyman factorization. Is this correct? If I'm on the right track, can someone provide a better explanation?