Showing that $f_\varphi(x)$ is a member of the one-parameter exponential family and $\sum_{i = 1}^n - \log(X_i)$ is sufficient for $\varphi$ Let $X_1, \dots, X_n$ denote a random sample from the PDF
$$f_{\varphi}(x)=
\begin{cases}
 \varphi x^{\varphi - 1} &\text{if}\, 0 < x < 1, \varphi > 0\\
      0 &\text{otherwise}
\end{cases}$$
I want to show that $f_\varphi(x)$ is a member of the one-parameter exponential family. Furthermore, I want to show that $\sum_{i = 1}^n - \log(X_i)$ is sufficient for $\varphi$.
Chapter 9.13.3 Exponential Families of the textbook All of Statistics: A Concise Course in Statistical Inference by Larry Wasserman says the following:

Most of the parametric models we have studied so far are special cases of a general class of models called exponential families. We say that $\{ f(x; \theta) : \theta \in \Theta \}$ is a one-parameter exponential family if there are functions $\eta(\theta)$, $B(\theta)$, $T(x)$ and $h(x)$ such that
$$f(x; \theta) = h(x) e^{\eta(\theta) T(x) - B(\theta)}.$$
It is easy to see that $T(x)$ is sufficient. We call $T$ the natural sufficient statistic.

I calculate the likelihood to be
$$\begin{align} L(\varphi; \mathbf{x}) &= \prod_{i = 1}^n \varphi x_i^{\varphi - 1} \mathbb{1}_{0 < x < 1, \varphi > 0} \\ &= \varphi^n x^{\sum_{i = 1}^n (\varphi - 1)} \prod_{i = 1}^n \mathbb{1}_{0 < x_i < 1, \varphi > 0} \\ &= \varphi^n x^{\sum_{i = 1}^n (\varphi - 1)} \mathbb{1}_{\text{min}(x_i) > 0, \varphi > 0} \end{align}$$
To get this in the appropriate form, I tried
$$\begin{align} \log\left[\varphi^n x^{\sum_{i = 1}^n (\varphi - 1)} \mathbb{1}_{\text{min}(x_i) > 0, \varphi > 0} \right] &= n\log(\varphi) + \sum_{i = 1}^n (\varphi - 1) \log(x_i) + \log \left[ \mathbb{1}_{\text{min}(x_i) > 0, \varphi > 0} \right] \\&= \exp{ \left\{ n\log(\varphi) + (\varphi - 1) \sum_{i = 1}^n \log(x_i) + \log \left[ \mathbb{1}_{\text{min}(x_i) > 0, \varphi > 0} \right] \right\} } \\ &= \exp{ \left\{ n\log(\varphi) + (\varphi - 1) \sum_{i = 1}^n \log(x_i) \right\} } \exp{\left\{ \log \left[ \mathbb{1}_{\text{min}(x_i) > 0, \varphi > 0} \right] \right\}} \\ &= \exp{ \left\{ n\log(\varphi) + (\varphi - 1) \sum_{i = 1}^n \log(x_i) \right\} } \mathbb{1}_{\text{min}(x_i) > 0, \varphi > 0} \end{align}$$
And we then select
$$\begin{align} &\eta(\varphi) = ? \\  &T(\mathbf{x}) = ? \\ &B(\varphi) = ? \\ &h(\mathbf{x}) = ? \end{align}$$
I'm unsure how much of this is correct. The first thing that might be problematic is $\log \left[ \mathbb{1}_{\text{min}(x_i) > 0, \varphi > 0} \right]$. Specifically, I'm not sure that it's valid to log this identity matrix in the first place, but I'm not sure what else to do. As you can see, I later use an exponential to cancel out the log, but, as I said, I'm not sure that it was mathematically valid to apply log to it in the first place. The second is that I needed to show that $\sum_{i = 1}^n - \log(X_i)$ is sufficient for $\varphi$, but I'm not sure how to further factorize the final expression in order to get this.
Another issue is that I cannot find any definitions or theorems for this in this chapter. There's usually a theorem that formally and explicitly states what you need to do/show in order show something (in this case, showing that $f_\varphi(x)$ is a member of the one-parameter exponential family and that $\sum_{i = 1}^n - \log(X_i)$ is sufficient for $\varphi$). The excerpt that I posted above was simply part of the normal discussion of the chapter at the very beginning. Based on that, I presumed that, for showing that $f_\varphi(x)$ is a member of the one-parameter family, I just have to show that the likelihood can take the form $f(x; \theta) = h(x) e^{\eta(\theta) T(x) - B(\theta)}$. However, for the sufficient statistic part, it just says that "it is easy to see that $T(x)$ is sufficient", so I'm not exactly sure what I'm supposed to do with this (do I need to use the Fisher-Neyman factorization theorem, or does the exponential family form imply that $T(\mathbf{x})$ is sufficient, and so we don't need to do anything else?).
So, overall, what is the correct way to do this to show that $f_\varphi(x)$ is a member of the one-parameter exponential family and $\sum_{i = 1}^n - \log(X_i)$ is sufficient for $\varphi$?
 A: I think I can give a short concise answer.
A pdf/pmf belongs to exponential family if you can write it in the form $$c(\theta)h(x)\exp(\sum_{i=1}^{k}w_i(\theta)t_i(x))$$
Further if $\theta=(\theta_1,\theta_2,...,\theta_d)$ and $d\leq k$ then $(\sum_{j=1}^{n}t_1(X_j), \sum_{j=1}^{n}t_2(X_j),...,\sum_{j=1}^{n}t_k(X_j))$ is sufficient for $\theta=(\theta_1,\theta_2,...,\theta_d)$.
Now you have the pdf  $$\varphi x^{\varphi-1}$$ where $\varphi>0$ and $0<x<1$ hence in this case $\theta\equiv \varphi$ which can be written as $$\varphi\exp((\varphi-1)\log x)$$ where $c(\theta)\equiv\theta\equiv \varphi$, $h(x)\equiv 1$, $k=1$, $w_1(\theta)\equiv w_1(\varphi)\equiv(\varphi-1)$ and $t_1(x)=\log x$. Hence it belongs to one-parameter exponential family. Also here $d=1=k$ therefore $\sum_{i=1}^{n}\log X_i$ is sufficient for $\varphi$ and any one-to-one function of sufficient statistic is also sufficient hence $\sum_{i=1}^{n}-\log (X_i)$ is also sufficient.
A: Our likelihood is $$L(\varphi;\textbf x)=\varphi^n\prod_{i=1}^n x_i^{\varphi-1}\mathbb 1_{\min (x_i)>0}\mathbb 1_{\max (x_i)<1}$$
Notice that we not only need the xi's to be at least 0, but the greatest of them must be no more than $1$.
This can be written as the following, because exponentiating and then taking the log is the same as the original expression. Don't apply the exponentiation-logarithm to the whole expression but rather to the specific part.$$\begin{align}L(\varphi;\textbf x)&=\mathbb 1_{\min (x_i)>0}\mathbb 1_{\max (x_i)<1}\exp\left\{\log \left(\varphi ^n\prod_{i=1}^n x_i^{\varphi-1}\right)\right\}\\
&=\mathbb 1_{\min (x_i)>0}\mathbb 1_{\max (x_i)<1}\exp\left\{n\log \varphi+\sum_{i=1}^n \log x_i^{\varphi-1}\right\}\\
&=\mathbb 1_{\min (x_i)>0}\mathbb 1_{\max (x_i)<1}\exp\left\{n\log \varphi+(\varphi - 1)\sum_{i=1}^n \log x_i\right\}\end{align}$$
We have $$\begin{align}h(x)&=\mathbb 1_{\min (x_i)>0}\mathbb 1_{\max (x_i)<1}\\
B(\varphi)&=-n\log \varphi\\
T(x)&=\sum_i \log x_i\\
\eta(\varphi)&=\varphi -1\end{align}$$
so $\sum_i \log x_i$ is a sufficient statistic by the factorization theorem. Note that the exponential family likelihood directly shows that $T(x)$ is the sufficient statistic as a result of the factorization theorem. So yes the form that the exponential family density is in allows you to conclude that $T(\textbf x)$ is the sufficient statistic, because $\varphi$ depends on $\textbf x$ only through $T(\textbf x)$. Also some linear transformation of a sufficient statistic is still sufficient so $-\sum_i \log x_i$ is sufficient.
At this point you could leave it at that or further simplify
$$\begin{align}L(\varphi;\textbf x)&=\mathbb 1_{\min (x_i)>0}\mathbb 1_{\max (x_i)<1}\exp\left\{n\log \varphi+\varphi \sum_{i=1}^n \log x_i-\sum_{i=1}^n \log x_i\right\}\\
&=\mathbb 1_{\min (x_i)>0}\mathbb 1_{\max (x_i)<1}\exp\left\{-\sum_{i=1}^n \log x_i\right\}\exp\left\{n\log \varphi+\varphi \sum_{i=1}^n \log x_i\right\}\end{align}$$
with
$$\begin{align}h(x)&=\mathbb 1_{\min (x_i)>0}\mathbb 1_{\max (x_i)<1}\exp\left\{-\sum_{i=1}^n \log x_i\right\}\\
B(\varphi)&=-n\log \varphi\\
T(x)&=\sum_i \log x_i\\
\eta(\varphi)&=\varphi \end{align}$$
so as you can see there is more than one way to factorize (the difference being $\eta(\varphi)$), however the conclusion remains the same.
In any case writing the likelihood in these forms shows that the pdf is part of the one-parameter exponential family based on the theorem cited.
