It seems like most of the times when we talk about product measures $\mathbf{R}^n$, we are talking about Lebesgue measure on $\mathbf{R}^n$. I will denote this measure space as $(\mathbf{R}^n, \mathcal{L}(\mathbf{R}^n), m_{\mathbf{R}^n})$.
On the other hand, one may consider the canonical product measure on $\mathbf{R}^n$ defined as product of the measure spaces $(\mathbf{R}, \mathcal{L}(\mathbf{R}), m_{\mathbf{R}})$, namely, the measure space $(\mathbf{R}^n, \otimes_n \mathcal{L}(\mathbf{R}), \pi)$, where $\pi: \otimes_{n} \mathcal{L}(\mathbf{R}) \to [0, +\infty]$ is the product measure extended by the Catheodory Extension Theorem from the premeasure $\pi_0: \{ \bigcup_{finite} A_i \times B_i: A_i, B_i \in \mathcal{L}(\mathbf{R}) \} \to [0, +\infty]$.
At the beginning I thought it would make sense that we have $(\mathbf{R}^n, \otimes_n \mathcal{L}(\mathbf{R}), \pi) = (\mathbf{R}^n, \mathcal{L}(\mathbf{R}^n), m_{\mathbf{R}^n})$. However, I have not yet succeeded in proving the equality. In particular, I am stuck at showing $$ \otimes_n \mathcal{L}(\mathbf{R}) = \mathcal{L}(\mathbf{R}^n). $$ We can consider this in the case where $n = 2$ and conclude by induction. Now I believe it is true that $$ \mathcal{L}(\mathbf{R}) \otimes \mathcal{L}(\mathbf{R}) \subseteq \mathcal{L}(\mathbf{R^2}) $$ since the cartesian product of two Lebesgue measurable sets are measurable: Is the product of two measurable subsets of $R^d$ measurable in $R^{2d}$?. I am not so sure about the converse inclusion: $$ \mathcal{L}(\mathbf{R}^2) \subseteq \mathcal{L}(\mathbf{R}) \otimes \mathcal{L}(\mathbf{R}) $$ Perhaps we need to argue through Borel $\sigma$-algebra and talk about the completion of measure spaces? Or is there a $\lambda$-$\pi$ system argument here? In fact, I fear this might not even be true. If this is not true, why do we introduce product measure as it is? What would be the motivation?
The inclusion $\mathcal{L}(\mathbb{R}^2) \subseteq \mathcal{L}(\mathbb{R}) \otimes \mathcal{L}(\mathbb{R})$ isn't true. What is true is that $\mathcal{L}(\mathbb{R}^2)$ is the completion of $\mathcal{L}(\mathbb{R})\otimes\mathcal{L}(\mathbb{R})$. Typically the product of two measure spaces will not be complete, even if the underlying measures are (see Product of complete measures). Of course, you can always take the completion, if you wish. Alternatively, you can restrict to the Borel sets, since $\mathcal{B}(\mathbb{R}^2) = \mathcal{B}(\mathbb{R}) \otimes \mathcal{B}(\mathbb{R})$.