My problem:
Let $X_1, X_2, ...,X_n$ be a random sample from the probability density function $$f\left(x;\theta\right)=\begin{cases} \theta \left(1-x\right)^{\theta-1} &: 0 <x_{i} < 1, 0< \theta \\ 0 &: \text{else} \end{cases}$$ Find the maximum likelihood estimator of $\theta$ and show that it is also a sufficient statistic.
My work so far:
For this probability density function I have found the M.L.E to be
$L\left(X_1,X_2, ...,X_n;\theta\right)= \prod^{n}_{i=1}f\left(x_{i};\theta\right)=\prod^{n}_{i=1}\left(\theta\left(1-x_{i}\right)^{\theta - 1}\right)=\theta^{n}\left(\prod^{n}_{i=1}\left(1-x_{i}\right)\right)^{\theta-1}$
Then, $\ln L= n \ln \theta + \left(\theta-1\right) \ln\left(\prod^{n}_{i=1}\left(1-x_{i}\right)\right)=n \ln \theta + \left(\theta-1\right)\sum^{n}_{i=1}\ln\left(1-x_{i}\right)$
Now, solving the partial derivative for $0$
$$\frac{\partial \ln L}{\partial \theta}=0 $$ $$\qquad = \frac{n}{\theta}+\sum^{n}_{i=1}\ln\left(1-x_{i}\right)=0$$
$$\implies \frac{n}{\theta}=-\sum^{n}_{i=1}\ln \left(1-x_{i}\right) $$
$$ \implies \theta = -\frac{n}{\sum^{n}_{i=1}\ln \left(1-x_{i}\right)} $$
So, the M.L.E. of $\theta$ is $\hat{\theta}=-\frac{n}{\sum^{n}_{i=1}\ln \left(1-x_{i}\right)}$.
From here I'm not sure how to show that this is a sufficient statistic. Should I use the factorization criterion? Any feedback is greatly appreciated.
You can absolutely use the Factorization theorem. In fact, you already have the factorization at the end of the first line of your MLE calculation: $$L(\theta \mid \boldsymbol x) = \theta^n \left(\prod_{i=1}^n (1-x_i)\right)^{\theta-1}$$ implies the choice $$h(\boldsymbol x) = 1, \quad T(\boldsymbol x) = \prod_{i=1}^n (1-x_i), \quad g(T \mid \theta) = \theta^n T^{\theta-1}.$$ Thus $T(\boldsymbol x)$ is sufficient for $\theta$, and the MLE $$\hat \theta = -\frac{n}{\log T},$$ which is a monotone function of $T$, is also sufficient for $\theta$.
I should note that to be fully correct, the likelihood function should actually be written $$L(\theta \mid \boldsymbol x) = \theta^n \left(\prod_{i=1}^n (1-x_i)\right)^{\theta-1} \prod_{i=1}^n \mathbb 1(0 < x_i < 1) = \theta^n \left(\prod_{i=1}^n (1-x_i)\right)^{\theta-1} \mathbb 1(0 < x_{(1)} \le x_{(n)} < 1),$$ hence the actual form of the factorization has $h(\boldsymbol x) = \mathbb 1(0 < x_{(1)} \le x_{(n)} < 1)$. This does not affect your MLE or proof of sufficiency, but because we require that the sample observations be contained in the interval $(0,1)$, the product of indicator functions and the subsequent minimum and maximum order statistics, are ancillary statistics for $\theta$.