Joint density function of i.i.d. r.v.’s form uniform distribution on $[0, \theta]$ Example
Suppose that $X_1,...,X_n$ are i.i.d. random variables with density function
$$f(x;\theta)=\begin{cases}
 \frac{1}{\theta}\quad \text{for all}\quad 0\le x\le \theta&  \\
 0 \quad \text{otherwise}
\end{cases}$$
where $\theta>0.$ The joint density function of $\mathbf{X}=(X_1,...,X_n)$ is
$$f(\mathbf{x};\theta)=\frac{1}{\theta^{n}}\mathbf{I}(\underset{1\le i\le n}{\text{max}} x_i\le\theta).$$

I am really puzzled why $f(\mathbf{x};\theta)=\frac{1}{\theta^{n}}\mathbf{I}(\underset{1\le i\le n}{\text{max}} x_i\le\theta)$, I think it should be $f(\mathbf{x};\theta)=\frac{1}{\theta^{n}}\mathbf{I}(\underset{1\le i\le n}{\text{max}} x_i\le\theta)\cdot \mathbf{I}(\underset{1\le i\le n}{\text{min}} x_i\ge 0)$.
If we restrict $X_{i}(1\le i\le n)$ are non-negative random variables,we can say that$$\frac{1}{\theta^{n}}\mathbf{I}(\underset{1\le i\le n}{\text{max}} x_i\le\theta)=\frac{1}{\theta^{n}}\mathbf{I}(\underset{1\le i\le n}{\text{max}} x_i\le\theta)\cdot \mathbf{I}(\underset{1\le i\le n}{\text{min}} x_i\ge 0).$$

Your own suggestion for the joint distribution is the correct one:
$$\begin{align}
f(\mathbf{x} | \theta) 
&= \prod_{i=1}^n \text{U}(x_i|0,\theta) \\[6pt]
&= \prod_{i=1}^n \frac{\mathbb{I}(0 \leqslant x_i \leqslant \theta)}{\theta} \\[6pt]
&= \frac{1}{\theta^n} \prod_{i=1}^n \mathbb{I}(0 \leqslant x_i \leqslant \theta) \\[6pt]
&= \frac{1}{\theta^n} \cdot \mathbb{I}(0 \leqslant x_{(1)} \leqslant x_{(n)} \leqslant \theta). \\[6pt]
\end{align}$$
Bear in mind that when we use the joint distribution to take the likelihood function we then get:
$$L_\mathbf{x}(\theta)
= \frac{1}{\theta^n} \cdot \mathbb{I}(\theta \geqslant x_{(n)}).
$$
