Let $N_1, N_2,\ldots N_k$ be normal subgroups of a group $G$ with $G = N_1N_2 \ldots N_k$. Assume for any $1 \leq i \leq k$, $N_i \cap (N_1N_2 \ldots N_{i-1}N_{i+1} \ldots N_{k}) = \langle e \rangle$. Define $N'_i = N_1N_2 \ldots N_i$. Is it true that $N_i \cap N'_{i-1} = \langle e \rangle$ for all $1 \leq i \leq k$? I want to show this as part of a larger proof to show $G$ is the direct product of the $N_i$s. I have already concluded that each $N'_i$ is normal, and know how to use induction to show my result if I have $N_i \cap N'_{i-1} = \langle e \rangle$.
I noticed that if there is an $a \neq e$ with $a \in N_i \cap N'_{i-1}$, $a \in N'_i$ but don't know how to proceed, or even if that is the correct direction.
I might also be overcomplicating things. It's possible I could show $G$ is the direct product of the $N_i$s if I show each element of $G$ has a unique representation in $N'_k$.
Take $N_i$ with $1 < i \leq k$. Let $a \in N_i \cap N'_{i-1}$. Note $a = a * e * e * \ldots e$. We can see $a * e * e * \ldots e \in N'_{i-1} N_{i+1} \ldots N_k$. Then, $a \in N_i \cap (N_1N_2 \ldots N_{i-1}N_{i+1} \ldots N_{k}) = \langle e \rangle$, so $a = e$ and $N_i \cap N'_i = \langle e \rangle$.