Given is the following distribution: $f_\theta(x)=\frac{1}{\theta}$ if $0<x\leq\theta$, and $0$ otherwise; $\theta<0$. I need to show that the maximum likelihood estimator of $\theta$, $\hat{\theta}=X_{n:n}$, is MSE consistent. The MSE is given by $Var(\hat{\theta})+(E[\hat{\theta}]-\theta)^2$. Thus I want to prove that this tends to $0$ as $n\rightarrow\infty$.
To calculate the first and second moment of $X_{n:n}$, I first determined the CDF: $(\frac{x}{\theta})^n$. By taking the derivative I found the density: $\frac{nx^{n-1}}{\theta^n}$.
So $E[X_{n:n}]=\int_{-\infty}^\infty \frac{xnx^{n-1}}{\theta^n}dx=\frac{n}{\theta^n}\cdot\int_{-\infty}^\infty x^ndx= ...$
If my previous calculations are correct, how can I calculate this integral?