Find the Maximum Likelihood Estimator

Let $X_1,X_2 ,X_3 ,..., X_n$ be iid with cdf $𝐹(𝑥) = 1 − 𝑥^{−𝜃}, 𝑥 > 0, 𝜃 > 0$.

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?

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 $$\operatorname{E}[\hat \theta] = \frac{n\theta}{n-1}, \quad n > 1.$$