Suppose that random variables $$X_{i1},...,X_{in}$$ are iid $\text{Uniform}(0,\theta_i),i=1,2.$Also, let $X_{1j}$'s be independent of $X_{2j}$'s. We assume that both parameters are unknown and $\theta=(\theta_1,\theta_2)\in\mathbb{R}^+\times\mathbb{R}^+$. Let us denote
$$X^{(n)}_i=\max_{1\leq j\leq n}X_{ij}.$$ Why the distribution function of $$X_i^{(n)}/\theta_i$$ is simply $t^n$ for $0<t<1$ and zero otherwise?
The subscript $i$ doesn't play a role here, so we can work with a single sample of $n$ iid uniform$(0,\theta)$ variables. Write $X^{(n)}$ for the max of these variables. Then $X^{(n)}/\theta$ is less than $t$ if and only if each of the $X_j/\theta$ is less than $t$. This allows us to compute the CDF of $X^{(n)}/\theta$: $$ P(X^{(n)}/\theta\le t)=P\left(\bigcap _{j=1}^n \left\{X_j/\theta\le t\right\}\right)=\left[P(X_1/\theta\le t)\right]^n $$ For $t\in(0,1)$ this last quantity is $t^n$, since each $X_j/\theta$ has uniform$(0,1)$ distribution. For $t\le0$ the requested probability is zero; for $t\ge1$ it's one.
