Given measurable spaces $(X,\mathcal{M})$ and $(Y,\mathcal{N})$, we can consider the product measure space $(X\times Y,\mathcal{M}\times\mathcal{N})$, where $\mathcal{M}\times\mathcal{N}$ is the $\sigma$-algebra generated by products $M\times N$, where $M\in\mathcal{M}$ and $N\in\mathcal{N}$ (this coincides with how you defined the product $\sigma$-algebra, and I think it is easier to see it this way).
If we have measure spaces $(X,\mathcal{M},\mu)$ and $(Y,\mathcal{N},\eta)$, then we can induce a measure $\mu\times\eta$, called the product measure, on $\mathcal{M}\times\mathcal{N}$ with the property that $(\mu\times\eta)(M\times N)=\mu(M)\cdot\eta(N)$ for $M\in\mathcal{M}$ and $N\in\mathcal{N}$. If the spaces in question are $\sigma$-finite, then the measure $\mu\times\eta$ with this property is unique.
Clearly, all this is true if we consider products of a finite number of measure spaces.
Given a topological space $X$, let $\mathscr{B}_X$ denote the Borel $\sigma$-algebra on $X$ (which is generated by open sets). If $X$ is second countable (i.e., if it has a countable topological basis), then $\mathscr{B}_{X^n}=\mathscr{B}_X\times\cdots\times\mathscr{B}_X=:(\mathscr{B}_X)^n$ (you can easily verify this). In particular, $\mathscr{B}_{\mathbb{R}^n}=(\mathscr{B}_\mathbb{R})^n$.
The usual definition of Lebesgue measure on $\mathbb{R}^n$ is the following: There exists a unique measure $\lambda$ on $\mathscr{B}_{\mathbb{R}^n}$ such that $\lambda_n(I_1\times\cdots I_n)=\operatorname{length}(I_1)\cdots\operatorname{length}(I_n)$ for all intervals $I_1,\ldots,I_n$. The Lebesgue measure on $\mathbb{R}^n$ is the completion of this measure, and it is also denoted by $\lambda$, and it is then defined in some other $\sigma$-algebra $\mathscr{L}_{\mathbb{R}^n}$ which contains $\mathscr{B}_{\mathbb{R}^n}$. When restricted to Borel sets, we have $\lambda_n=\lambda_1\times\cdots\times\lambda_1$.
However, it is not true that $\mathscr{L}_{\mathbb{R}^{n+m}}=\mathscr{L}_{\mathbb{R}^n}\times\mathscr{L}_{\mathbb{R}^n}$ (to be precise, this is not true if we assume the Axiom of Choice).
Finally, to solve the question, you should first show that $\mathscr{L}_{\mathbb{R}^n}\times\mathscr{L}_{\mathbb{R}^m}\subseteq\mathscr{L}_{\mathbb{R}^{n+m}}$. You can do this using the definition of completion of measures, and using the fact that $\lambda_{n+m}=\lambda_n\times\lambda_m$ on Borel sets. Since $\mathscr{L}_{\mathbb{R}^{n+m}}$ is a complete $\sigma$-algebra (with respect to $\lambda_n$), then it contains the completion of $\mathscr{L}_{\mathbb{R}^n}\times\mathscr{L}_{\mathbb{R}^m}$.
On the other hand, it is clear that $\mathscr{L}_{\mathbb{R}^n}\times\mathscr{L}_{\mathbb{R}^m}$ contains $\mathscr{B}_{\mathbb{R}^n}\times\mathscr{B}_{\mathbb{R}^m}=\mathscr{B}_{\mathbb{R}^{n+m}}$, so the completion of $\mathscr{L}_{\mathbb{R}^n}\times\mathscr{L}_{\mathbb{R}^m}$ contains the completion of $\mathscr{B}_{\mathbb{R}^{n+m}}$, which is by definition $\mathscr{L}_{\mathbb{R}^{n+m}}$ (here we again use the fact that $\lambda_{n+m}=\lambda_n\times\lambda_m$).