Finite abelian group generated by elements of maximal order I'm trying to prove that a finite abelian group $G$ is generated by elements of maximal order.

I can sort of see why that happens in a vague instinctive kind of way but no real hard logic. So far, I have tried to use this lemma:

Let $G$ be a finite abelian group and a be an element of maximal order in $G$. Then any element $b$ is such that $|b|$ divides $|a|$.

Then each cyclic group generated by the maximal order element contains exactly 1 element of each possible order for an element in the group. I have tried expanding from this idea in a few directions but I don't think it's the correct way to go since it's not going anywhere.

Any help appreciated. Thanks guys.
Let $G$ be a finite abelian group, then $G = \mathbb{Z}_{n_1}\oplus\cdots\oplus \mathbb{Z}_{n_k}$ for some $n_1|\cdots|n_k$. Then clearly these elements 
$$\{(1,\cdots,1),(0,1,\cdots,1),(0,0,1,\cdots,1),\cdots,(0,\cdots,0,1)\}$$
are of maximal order $n_k$ and they generate $G$.
let a be an element of a prime order such that is not belong to the subgroup of G generated by all maximal elements then ab is not belong to the set of the generators of maximal elemens for some b of maximal elements.