Typical proof for this goes somewhat like this
Consider a, $a^2$, $a^3$ etc. Since there are only finitely many elements we must have $a^m = a^n$ for some m < n.
I understand that because it's finite group at some point $a^n$ will not keep increasing but will go lower. But I don't get why $a^m$ has to equal to $a^n$.
Can't $a^m = b$ where b < $a^m$ but is a different element which is not a power of a. Won't this still satisfy the condition of finiteness of the group?