Suppose $(X_1,...,X_n)$~$P_{n,\theta}=\mathscr{U}([0,\theta])^n$ (Uniform distribution). Let $Z_n=n(\theta_0-X_{(n)})$ where $X_{(n)}=\max(X_1,...,X_n)$, and let $Z$~$\mathscr{E} \left(\frac{1}{\theta_0} \right)$ (Exponential distribution). How do I show that $Z_n$ converges in distribution to $Z$ under $P_{n,\theta_0}$?
For convergence in probability, I need to show the CDF of $Z_n$ goes to the CDF of $Z$ as $n$ goes to infinity. I found the CDF of $Z$ to be $F_Z(x)=\frac{x-0}{\theta_0-0}=\frac{x}{\theta_0}$, but how can I find the CDF of $Z_n$?
Hint: Argue that $\theta_0-X_{(n)}=\min\left\{\theta_0-X_1,\theta_0-X_2,\ldots,\theta_0-X_n\right\}$. And the min of a list of numbers exceeds threshold $t$ iff every number exceeds $t$. Therefore the survival function for $Z_n$ is $$P(Z_n>t):=P\left(\theta_0-X_{(n)}>\frac tn\right)=P\left(\bigcap_{i=1}^n \left\{\theta_0-X_i >\frac tn\right\}\right).%=\prod_{i=1}^nP\left(\theta_0-X_i>\frac tn\right). $$ Now argue that the RHS converges to $P(Z>t)$.
You have the wrong cdf for $Z$. The limiting distribution is not uniform over $[0,\theta_0]$. It's exponential, with survival function $P(Z>t)=\exp(-t/\theta_0)$.