Prove that if $G$ has an exponent $n$ then the order of $G$ divides some power of $n$. 
Prove that if $G$ has an exponent $n$ then the order of $G$ divides some power of $n$.

I found the proof in Lang's Algebra on Page 33, revised Third edition,  section 6 of the first chapter. I'll mention briefly the parts I understood and the parts that I need help with.
It's a proof by induction, apparently. We first take a cyclic group generated by some element $b$ of $G$ and then show that the order of the cyclic group generated divides $n$. However, Lang then says that the order of $G/H$ divides a power of $n$ by induction.
My question is, what induction?
I think I get the rest of the proof, any help with this part will be highly appreciated. Thanks!

This is part of the proof of Lemma 6.1, in Section I.6 of Lang's Algebra, revised 3rd Edition, Springer-Verlag.
Though you "forgot" to mention it, there is an assumption that $G$ is finite abelian. This is incredibly important for the proof you outline, since there is an immediate assumption that you can mod out by the subgroup $H$, even though it was never shown to be normal. Please try to include information that is so relevant that the proof makes no sense without it.
The induction is on the order of the group. Fix $n$. Let $G$ be a finite abelian group of exponent $n$, and we wish to show that the order of $G$ divides some power of $n$. Assume by induction that:

If $A$ is any abelian group of order strictly smaller than $|G|$, and $A$ has exponent $n$, then the order of $A$ divides some power of $n$.

That is, we are doing strong induction on the order of $G$.
Let $b\in G$, $b\neq 1$, and let $H=\langle b\rangle$. The order of $H$ divides $n$, because $b^n=1$ (since $G$ has exponent $n$). Now let $A=G/H$. This is a group of order strictly smaller than $|G|$ (it has order $|G|/|H|$ and $|H|\gt 1$), and has exponent $n$ (any quotient of a group of exponent $n$ has exponent $n$). So by the induction hypothesis, $|A|$ divides a power of $n$; say $|A|$ divides $n^r$. Then $|G|/|H|$ divides $n^r$, so $|G|$ divides $|H|n^r$. But since $|H|$ divides $n$, then $|H|n^r$ divides $nn^r=n^{r+1}$. Thus, $|G|$ divides $n^{r+1}$, proving the result for $G$.
The strong induction argument assumes that $G$ has nontrivial elements, so we need to prove a special case: if $G$ has no nontrivial elements, then $|G|=1$, which of course divides $n$, a power of $n$.
