Explicitly show that the product Lebesgue measure satisfies $m^1 \otimes m^1 = m^2$

Question

From Caratheodory's uniqueness theorem it can be shown that the product topology on $\mathbb{R}^2$ satisfies $m^1 \otimes m^1 = m^2$. I am trying to show a weaker version using only the definition of the outer measure induced from $m^1 \otimes m^1$ and $m^2$ respectively, that is: $$m^1(E)m^1(F) = m^2(E\times F)$$ using only the definition $$(m^2)^*(E)=\inf \{\sum_{n=1}^\infty Vol(R_n)\ |\ E\subset\bigcup_{n=1}^\infty R_n,\ R_n=[a_n,b_n]\times[c_n,d_n]\}\ $$ for every Borel sets E and F in $\mathbb{R}$. It is immediate that $$m^2(E\times F) \leq m^1(F)m^1(E)$$ but i can't find a way to prove the second inequality. Any hints or different directions will be welcome.