Since the semigroup \(S\) is finite, for any \(x \in S\), the set of its powers \(\{ x^1, x^2, x^3, \dots \} \) must also be finite. By the pigeonhole principle, there must exist distinct positive integers \(m, n\) such that \(x^m = x^n\). Let’s assume \(m {\lt} n\). Then we can write \(x^m = x^m x^{n-m}\). This shows that powers of \(x\) eventually enter a cycle. From this cyclic part, we can extract a power \(k\) such that \(x^k\) is idempotent. The proof of uniqueness follows by showing that if \(x^a\) and \(x^b\) are both idempotents, then \(x^a = x^b\).

From \(a = xay\), we can repeatedly substitute \(a\) into itself to get \(a = x^k a y^k\) for any \(k \geq 1\). Since \(M\) is a finite monoid, there exists a non-zero power \(n_1\) such that \(x^{n_1}\) is an idempotent (by 1). Then we have \(a = x^{n_1} a y^{n_1}\). Multiplying by \(x^{n_1}\) on the left gives \(x^{n_1}a = x^{2n_1} a y^{n_1}\). Since \(x^{n_1}\) is idempotent, \(x^{2n_1} = x^{n_1}\), so \(x^{n_1}a = x^{n_1} a y^{n_1} = a\). A symmetric argument shows that \(ay^{n_2} = a\) for some idempotent power \(y^{n_2}\).