Before proving the theorem we note a special case that is frequently used: Observe that for normal subgroups $N_1,N_2,…,N_r$ of a group $G$, $\langle N_1\cup N_2\cup \dotsb\cup N_r\rangle =N_1N_2\dotsb N_r=\{n_1n_2\dotsb n_r\mid n_i\in N_i\}$ by an easily proved generalization of theorem 3 section 5. In additive notation $N_1N_2\dotsb N_r$ is written $N_1+N_2+\dotsb +N_r$.
My attempt: We use mathematical induction approach. Base case: $i=2$. Since $N_2\lhd G$, we have $\langle N_1\cup N_2\rangle=N_1N_2$ by theorem 3 section 5. Inductive step: Suppose $\langle N_1\cup \dotsb\cup N_i\rangle =N_1\dotsb N_i$, for some $2\leq i\lt r$. By this post (put $G_1=G_2$), $\langle N_1\cup \dotsb\cup N_i\cup N_{i+1}\rangle =\langle \langle N_1\cup \dotsb \cup N_i\rangle \cup N_{i+1}\rangle.$ Since $N_{i+1}\lhd G$ and by inductive hypothesis, we have $$\langle \langle N_1\cup \dotsb \cup N_i\rangle \cup N_{i+1}\rangle = \langle N_1\cup \dotsb \cup N_i\rangle N_{i+1}=N_1\dotsb N_iN_{i+1}.$$ Thus $\langle N_1\cup \dotsb\cup N_i\cup N_{i+1}\rangle = N_1\dotsb N_iN_{i+1}$. By principle of mathematical induction, $\forall 2\leq i\leq r$ we have $\langle N_1\cup \dotsb\cup N_i\rangle = N_1\dotsb N_i$. Hence $\langle N_1\cup \dotsb\cup N_r\rangle = N_1\dotsb N_r$. Is my proof correct?
In above proof we didn’t use $N_1\lhd G$ condition, did we? I think it is enough to assume either $N_1\lhd G$ or $N_2\lhd G$.