Show that $\frac1n(\sum_{i=1}^n\log\frac{1}{1-X_i})^3$ is a sufficient statistic for $\beta$ in a Beta$(\alpha,\beta)$ density Suppose that $X_1,...,X_n$ From a random sample from a distribution with density $$ f(x) = \frac{\Gamma{(\alpha+\beta)}}{\Gamma{(\alpha)}\Gamma{(\beta)}}x^{\alpha-1}(1-x)^{\beta-1} \quad,\text{ if }\space x \in(0,1)$$
For some $\alpha$ > 0 known, and $\beta > 0$ unknown. Show that
$$ T(X_1,...,X_n)=\frac1n\left(\sum_{i=1}^n\log\frac{1}{1-X_i}\right)^3$$
is a sufficient statistic for $\beta$.
SOLUTION:
I used Fisher Factorization and managed to show that $$S(x_1,\ldots,x_n)=\prod_{i=1}^n(1-x_i)$$
is sufficient for $\beta$.
However I do not now how to get to $T(X_1,...,X_n)=\frac1n\left(\sum_{i=1}^n\log\frac{1}{1-X_i}\right)^3$ from $S(X_1,\ldots,X_n)=\prod_{i=1}^n(1-X_i)$. 
 A: For $\alpha,\beta>0$, joint density of $(X_1,X_2,\cdots,X_n)$ is 
\begin{align}
f_{\beta}(x_1,x_2,\cdots,x_n)&=\frac{1}{\left(B(\alpha,\beta)\right)^n}\left(\prod_{i=1}^n x_i\right)^{\alpha-1}\left(\prod_{i=1}^n(1-x_i)\right)^{\beta-1}\mathbf1_{0<x_i<1}
\\&=\exp\left[-n\ln B(\alpha,\beta)+(\alpha-1)\sum \ln x_i+(\beta-1)\sum \ln(1-x_i)\right]
\\&=\exp\left[\beta\sum_{i=1}^n \ln(1-x_i)+A(\alpha,\beta)+B(x_1,\cdots,x_n)\right]
\end{align}
for some function $A$ and $B$.
Clearly, the family of distributions $\{f_{\beta}:\beta>0\}$ belongs to the one-parameter exponential family. Hence, a minimal sufficient statistic for $\beta$ is $$H(X_1,\cdots,X_n)=\sum_{i=1}^n\ln (1-X_i)$$
Observe that 
\begin{align}T(X_1,\cdots,X_n)&=\frac{1}{n}\left[\sum_{i=1}^n\ln\left(\frac{1}{1-X_i}\right)\right]^3
\\&=\frac{-1}{n}\left[\sum_{i=1}^n\ln(1-X_i)\right]^3
\end{align}
is a function of $H(X_1,\cdots,X_n)$. 
And we know that a minimal sufficient statistic is a function of every other sufficient statistic. 
So $T$ is a sufficient statistic for $\beta$.
We can also show that the joint density can be factored as $$f_{\beta}(x_1,\cdots,x_n)=g(\beta, T)h(x_1,\cdots,x_n)$$
where $g$ depends on $\beta$ and on $x_1,\cdots,x_n$ through $T$ and $h$ is independent of $\beta$.
You say you could show that $S=\prod_{i=1}^n (1-X_i)$ is sufficient for $\beta$. But then $T$ is a function of $S$ also. So we reach a similar conclusion from the Factorisation theorem.
A: $$-nT = (\sum\limits_{i=1}^n \log(1-x_i))^3$$
$$\sqrt[3]{-nT} = \log(\Pi_{i=1}^n (1-x_i))$$
$$\exp(\sqrt[3]{-nT}) = S$$
The key is that all transformations are bijective.
