Find the Maximum Likelihood Estimator Let $X_1,X_2 ,X_3 ,..., X_n$ be iid with cdf $() = 1 − ^{−},  > 0,  > 0$.
What I have done so far,
$L(\theta)=\prod\limits_{i=1}^n\theta x_i^{\theta -1}=\theta^n\prod\limits_{i=1}^n\ x_i^{\theta -1}$
$ln(L(\theta))=nln(\theta)+(\theta-1)^{n}\prod\limits_{i=1}^n ln(x_i)$
$\frac{d}{d\theta}ln(L(\theta))=\frac{n}{\theta}+n(\theta-1)^{n-1}\prod\limits_{i=1}^nln(x_i)=0$
$\theta(\theta -1)^{n-1}=\frac{-1}{\prod\limits_{i=1}^nln(x_i)}$
How do I proceed?
 A: You've got a number of errors.  First, the support should be $x > 1$, since $1 - x^{-\theta} < 0$ if $x < 1$.  Second, if $$F_X(x) = 1 - x^{-\theta}, \quad x > 1, \; \theta > 0,$$ then $$f_X(x) = F_X'(x) = \theta x^{-(\theta+1)}, \quad x > 1, \; \theta > 0$$ and the likelihood may be written as $$\mathcal L(\theta \mid \boldsymbol x) \propto \prod_{i=1}^n \theta x_i^{-(\theta+1)} \mathbb 1 (x_{(1)} > 1) \mathbb 1 (\theta > 0).$$  The log-likelihood is then $$\ell(\theta \mid \boldsymbol x) = n \log \theta - (\theta+1) \log \prod_{i=1}^n x_i + \log \left( \mathbb 1(x_{(1)} > 1) \mathbb 1 (\theta > 0) \right).$$  The derivative of the log-likelihood when the indicator functions are $1$ is $$\frac{\partial \ell}{\partial \theta} = \frac{n}{\theta} - \log \prod_{i=1}^n x_i,$$ which gives a critical point $$\hat \theta = \frac{n}{\log \prod_{i=1}^n x_i} = \frac{n}{\sum_{i=1}^n \log x_i} = \left( \overline{\log x_i} \right)^{-1};$$ that is to say, $\hat \theta$ is the reciprocal of the arithmetic mean of the log-transformed sample.  I leave it as an exercise to demonstrate this choice is maximal under the conditions $x_{(1)} > 1$ and $\theta > 0$.

To calculate the bias of $\hat \theta$, the nature of the MLE suggests that we should try to log-transform the distribution from which the sample is drawn.  That is to say, consider $Y = g(X) = \log X$.  Then $$f_Y(y) = f_X(g^{-1}(y)) \left|\frac{dg^{-1}}{dy}\right| = \theta (e^y)^{-(\theta+1)} e^y = \theta e^{-\theta y}, \quad y > 0,$$ since $x > 1$ implies $y = \log x  > 0$.  Thus we recognize $Y$ as an exponential distribution with rate $\theta$, and we can express the MLE in terms of the log-transformed sample $\boldsymbol y = \log \boldsymbol x$:  $$\hat \theta = \frac{1}{\bar y}.$$  The log-sample total $n\bar y$ is gamma distributed with shape $n$ and rate $\theta$:  $$f_{n \bar Y}(y) = \frac{\theta^n y^{n-1} e^{-\theta y}}{\Gamma(n)}, \quad y > 0, \theta > 0,$$ being the sum of IID exponential variables; thus $\hat \theta /n = 1/(n \bar y)$ is inverse gamma distributed:  $$f_{\hat \theta/n}(y) = \frac{\theta^n e^{-\theta/y}}{y^{n+1} \Gamma(n)}, \quad y > 0, \theta > 0.$$  I leave it as an exercise to show this has expectation $$\operatorname{E}[\hat \theta/n] = \frac{\theta}{n-1}, \quad n > 1,$$ hence $$\operatorname{E}[\hat \theta] = \frac{n\theta}{n-1}, \quad n > 1.$$
A: your problem was with logarithms 
$$
\ln\left(\prod_{i=0}^k x_i^a\right) = a\sum_{i=0}^k \ln\left(x_i\right)
$$
which makes your problem easier.
A: \begin{align*}
L(\theta)&=\theta^n\prod\limits_{i=1}^n\ x_i^{-\theta -1} \\
\implies \ell(\theta) &= n \ln(\theta) + \ \sum_{i=1}^n (-\theta -1) \ln(x_i)\\
\implies \frac{d \ell}{d \theta} &= \frac{n}{\theta} - \sum_{i=1}^n \ln(x_i)
\end{align*}
and checking where this equals zero gives $\hat \theta_{MLE} = n \ / \left( \sum_{i=1}^n \ln(x_i) \right) $
We also check $\frac{d^2 \ell}{d \theta^2} = - \frac{n}{\theta^2}< 0$ so this is indeed a maximium.
A: I've already correct that mistake, I hope the solution would be usefull. By the way, in the first part you can use the exponents law's to do $a^nb^n=(ab)^n$, and then use the logarithm law's.
