If $G$ is cyclic, then $G=\{a^0, a^1, ...\}$, but why does it have to be that $a^k=e$ for some $k$? If $G$ is cyclic, then $G=\{a^0, a^1, ...\}$, but why does it have to be that $a^k=e$ for some $k$?
I.e. that for $G$ to be cyclic, then surely the generating element would need to generate $e$ also.
What implications does this ($a^k=e$ for some $k$) have for the group elements?
 A: You're forgetting the assumption that $G$ is finite and that you want $k>0$ (otherwise the statement is trivially satisfied for $k=0$). Since $G$ is finite, there must exist integers $r$ and $s$ with $r>s>0$ and $a^r=a^s$.
Now, take into account that $a^{-s}\in G$ and multiply both sides by it.
A: You have $a^0=e$ if by $e$ you mean the neutral element of the group.
If your group $G$ is finite, take $k:=\vert G\vert$, and then $$a^k=e.$$
In some cases, it even exists a smaller $k$, which divides $\vert G\vert$ by Lagrange theorem.
If your groupe is not finite (take $\mathbb Z$), then you won't necessarily have a $k\ne 0$ such that $ak=e$ (it is an additive group so $a^k$ becomes $ak$). But you might with some groups.
A: If $a$ is a generator of the cyclic group $G$ then $$\phi:\quad({\mathbb Z},+)\to(G,*),\qquad k\mapsto a^k$$
is a surjective homomorphism. If $\phi$ is injective then $G$ is isomorphic to ${\mathbb Z}$. If $\phi$ is not injective then there are $j$, $k\in{\mathbb Z}$ with $j\ne k$ and $\phi(j)=\phi(k)=:u$. Put $m:=j-k\ne0$. Then $$a^m=a^{j-k}=a^j*\bigl(a^k\bigr)^{-1}=u*u^{-1}=e\ .$$
