A question on Finite Semigroup

Question

Why is it so, that a finite semigroup, say $(S, \circ)$ has $a^m=a^n$ for positive integers $m$ and $n$ with $m>n$ for $a\in S$? Does it imply some sort of periodicity in the binary composition?

Answer 1

If no two of $a^1, a^2, a^3, \ldots$ were the same, then the semigroup would not be finite.

Answer 2

You have that $a^{m-n}=e$ the identity, where $m-n>0$ is an integer. So that, there exist a minimal integer $k>0$ such that $a^k=e$. Then we have that the elements of the group are $\{a,a^2,\ldots,a^{k-1},a^k=e\}$.