proof that a finite abelian group has element with $69640972g86418188=exp(G)$

Theorem 1

If $G$ is a finite abelian group, then $G$ has an element of order $\exp \mathit{}\mathrm{(}G\mathrm{)}$.

Proof. Write $\exp (G)=\prod p_i^{k_i}$. Since $\exp (G)$ is the least common multiple of the orders of each group element, it follows that for each $i$, there is an element whose order is a multiple of $p_i^{k_i}$, say $|c_i|=a_i{p}_i^{k_i}$. Let $d_i=c_i^{a_i}$. Then $|d_i|=p_i^{k_i}$. The $d_i$ thus have pairwise relatively prime orders, and thus

$$\left|\prod d_i\right|=\prod \left|d_i\right|=\exp (G)$$

so that $\prod d_i$ is the desired element.