Proof. We prove by induction on $n$ For $n=2$ $I_1+I_2 = R$ implies:\begin{equation} I_1\cap I_2 = R(I_1\cap I_2) = (I_1 + I_2)(I_1 \cap I_2) \subseteq I_1I_2. \end{equation}The converse inclusion is trivial. Assume now that the equality holds for $n \ge 2$ $J:= I_1 \cap I_2 \cap ... \cap I_n = I_1I_2 ... I_n$ Since $ I_{n+1}+I_j = R$ for every $j \neq {n+1}$ there exist the elements $a_j\in I_j$ and $b_j\in I_{n+1}$ such that $a_j + b_j =1$ The product $c:= \prod_{j=1}^n a_j = \prod_{j=1}^n(1-b_j)\in 1 + I_{n+1}$ Also $c\in J$ then $1\in J+I_{n+1}$ or $J+I_{n+1} =R$
Applying the case $2$ the induction step is satisfied:\begin{equation} I_1I_2 ... I_{n+1} = J I_{n+1} = J\cap I_{n+1} = I_1\cap I_2 \cap ... \cap I_n \cap I_{n+1}. \end{equation}