If $G$ is a finite group and $N_1,\dots, N_n$ are normal subgroups of $G$ such that $G=N_1N_2\dots N_n$ and $o(G)=o(N_1)o(N_2)\dotso(N_n)$, prove that $G$ is the direct product of $N_1,\dots, N_n$.
I am trying to prove that $$N_i\cap \prod\limits_{j=1,\ j\neq i}^{n} N_j=\{e\}.$$ If we are able to show that then $G$ is direct product of $N_1,\dots, N_n$. $$o(N_1)o(N_2)\dotso(N_n)=o(N_1N_2\dots N_n)=\dfrac{o(N_1)o(N_2\dots N_n)}{o(N_1\cap N_2\dots N_n)}=$$$$=\dfrac{o(N_1)o(N_2)o(N_3\dots N_n)}{o(N_1\cap N_2\dots N_n)o(N_2\cap N_3\dots N_n)}=\dots =\dfrac{o(N_1)o(N_2)\dotso(N_n)}{o(N_1\cap N_2\dots N_n)o(N_2\cap N_3\dots N_n)\dots o(N_{n-1}\cap N_n)}.$$ So we get the following: $o(N_1\cap N_2\dots N_n)o(N_2\cap N_3\dots N_n)\dots o(N_{n-1}\cap N_n)=1$ and hence $$N_1\cap N_2\dots N_n=\{e\},\ N_2\cap N_3\dots N_n=\{e\},\dots, \ N_{n-1}\cap N_n=\{e\}.$$ How to show that for example $N_2\cap N_1N_3\dots N_n=\{e\}$ ?
You are basically done.
Observe that we can arbitrarily rearrange $N_i$'s, as they commute with each other, because they are normal: $$\forall g:gN_ig^{-1}=N_i\implies\forall g:gN_i=N_ig\implies N_jN_i=N_iN_j$$ so that we equally well can write $G=N_{\sigma(1)}N_{\sigma(2)}\dots N_{\sigma(n)}$ for any permutation $\sigma\in S_n$.
And by the way this is the property of normal subgroups that makes e.g. $N_2N_3\dots N_n$ equal to the smallest subgroup that contains each of $N_2,N_3\dots,N_n$. (We actually have to check $N_1\cap\langle N_2,\dots,N_n\rangle=\{e\}$.)