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) 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) If $$l > 1$$, then:

$$ $$ $$ $$ $$ $$

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