Proof. By the division algorithm, there exist $q,r \in {\mathbb Z}$ with $0 \le r<\operatorname{exp}~G$ such that $|G|=q(\operatorname{exp}~G)+r$ Let $g \in G$ Then $e_G=g^{|G|}=g^{q(\operatorname{exp}~G)+r}=g^{q(\operatorname{exp}~G)}g^r=(g^{\operatorname{exp}~G})^qg^r=(e_G)^qg^r=e_Gg^r=g^r$ Thus, for every $g \in G$ $g^r=e_G$ By the definition of exponent, $r$ cannot be positive. Thus, $r=0$ It follows that $\operatorname{exp}~G$ divides $|G|$