# written proof – Prove that if \$ {7k: k in {Z} } subsetneq {nm: m in {Z} } \$, then n = 1.

Let n be a natural number. Prove that if $${7k: k in {Z} } subsetneq {nm: m in {Z} }$$, then n = 1.

I know that we have to show $$x in {A}$$ involved $$x in {B}$$, and that there is $$x in {B}$$ such as $$x notin {A}$$. Which means that $$x in {A} neq {x in {B}}$$.
Any help is welcome, thank you!