Element has Idempotent Power in Finite Semigroup

Theorem
Let $\left({S, \circ}\right)$ be a finite semigroup.

For every element in $\left({S, \circ}\right)$, there is a power of that element which is idempotent.

That is:
 * $\forall x \in S: \exists i \in \N: x^i = x^i \circ x^i$

Proof
From Finite Semigroup Equal Elements for Different Powers, we have:

$\forall x \in S: \exists m, n \in \N: m \ne n: x^m = x^n$.

Let $m > n$. Let $n = k, m = k + l$.

Then $\forall x \in S: \exists k, l \in \N: x^k = x^{k + l}$.

Now we show that $x^k = x^{k + l} \implies x^{k l} = x^{k l} \circ x^{k l}$, i.e. that $x^{k l}$ is idempotent.

First:

From here we can easily prove by induction that $\forall n \in \N: x^k = x^{k + n l}$.

In particular, $x^k = x^{k + k l} = x^{k \left({l + 1}\right)}$.

There are two cases to consider:


 * $(1): \quad$ If $l = 1$, then $x^k = x^{k \left({l + 1}\right)} = x^{2 k} = x^k \circ x^k$, and $x^{k l} = x^k$ is idempotent.


 * $(2): \quad$ If $l > 1$, then:

... and again, $x^{k l}$ is idempotent.