Showing that estimator is consistent Let $\hat{\theta}_n= -\frac{n}{\sum_{i=1}^n \log(X_i)}$, where $X_i$ are i.i.d. samples from distribution with pdf $\theta x^{\theta-1}$ for $x \in (0,1)$. How to prove that $\hat{\theta}_n$ is consistent estimator of $\theta$.
 A: Probability limits distribute, so 
$$\text{plim}\left (\frac {1}{\hat \theta}\right) = \frac {1}{\text{plim}\hat \theta}$$
Also, 
$$\hat{\theta}_n= -\frac{n}{\sum_{i=1}^n \log(X_i)} \implies \frac {1}{\hat \theta} = -\frac 1n\sum_{i=1}^n \log(X_i) $$
The sample is i.i.d. and so ergodic and so 
$$\text{plim}\left (\frac {1}{\hat \theta}\right) = \text{plim} \left(-\frac 1n\sum_{i=1}^n \log(X_i)\right) = E\left[-\log(X_i)\right]$$
So you just have to find the distribution of $Y = -\log(X_i)$ as suggested in a comment, and calculate its expected value.
A: We can also use the sufficient condition of consistency showing that $E_{\theta}(\hat\theta_n)\to\theta$ and $\operatorname{Var}_{\theta}(\hat\theta_n)\to0$ as $n\to\infty$ to prove that $\hat\theta_n$ is consistent for $\theta$. But then again, one needs to know the distribution of the sufficient statistic $\sum_{i=1}^n\ln X_i$.
Since the population DF is of the form $F_{\theta}(x)=x^{\theta}$ for $0<x<1$, we have $X^{\theta}\sim\mathcal U(0,1)$. 
Thus,
\begin{align}
-\ln X^{\theta}&\sim\text{Exp}(1)
\\\text{ or },\,-\theta\ln X&\sim\text{Exp}(1)
\end{align}
Let $T=\sum_{i=1}^n\ln X_i$ where $-\theta \ln X_i$ are i.i.d $\text{Exp}(1)$ variables.
Then, $-\theta \,T\sim\mathcal{G}(1,n)$ with density $$g(t)=\frac{e^{-t}t^{n-1}}{\Gamma(n)}\mathbf1_{t>0}$$
The MLE of $\theta$ is $$\hat\theta_n=-\frac nT$$
Now,
\begin{align}
E(\hat\theta_n)&=n\theta E\left(\frac{1}{-\theta T}\right)
\\&=n\theta \int_0^\infty \frac{1}{t}\frac{e^{-t}t^{n-1}}{\Gamma(n)}\,dt
\\&=n\theta\,\frac{\Gamma(n-1)}{\Gamma(n)}
\\&=\frac{n\theta}{n-1}
\\&=\frac{\theta}{1-\frac{1}{n}}\longrightarrow\theta\quad\text{ as }n\to\infty
\end{align}
Similarly, 
\begin{align}E\left(\hat{\theta_n}^2\right)&=n^2\theta^2E\left(\frac{1}{\theta^2T^2}\right)
\\&=\frac{n^2\theta^2}{(n-1)(n-2)}
\end{align}
Therefore, 
\begin{align}
\operatorname{Var}(\hat\theta_n)&=E\left(\hat\theta_n^2\right)-\left(E(\hat\theta_n)\right)^2
\\&=\frac{n^2\theta^2}{(n-1)^2(n-2)}
\\&=\frac{\theta^2}{\left(1-\frac{1}{n}\right)^2(n-2)}\longrightarrow0\quad\text{ as }n\to\infty
\end{align}
Hence, $\hat\theta_n$ is a consistent estimator of $\theta$.
