Proof
(related to Proposition: Existence and Uniqueness of Greatest Elements in Subsets of Natural Numbers)
We will prove the proposition for the poset $(\mathbb N,\le).$ An analogous proof is possible for the strictlyordered set $(\mathbb N, < ),$ in which we have to replace the order relation "$\ge$" by the order relation "$>$" and the word maximum by the word maximal element.
 Let \(M\) be a nonempty and finite subset of natural numbers \(M\subseteq\mathbb N.\)
 Since $M$ is nonempty, let \(n_0\in M\) be its element.
 We will construct an maximum of $M$ as follows:
 If for all $m\in M$ we have $m\ge n_0$ we set \(m_0:=n_0\), since it is the maximum we were looking for.
 Otherwise there exists an $n_1\in M$ with \(n_1 > n_0.\)
 We ask wether for all $m\in M$ we have $m\ge n_1.$
 If so, then we set \(m_0:=n_1\), otherwise we find an $n_2\in M$ with \(n_2 > n_1.\) and repeat the same procedure again.
 Since $M$ is finite, our searching process will be repeated at most \(n_0  m_0+1\) times until we finally find the maximum of \(M\).
 By construction $m_0$ is unique.
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References
Bibliography
 Kramer Jürg, von Pippich, AnnaMaria: "Von den natürlichen Zahlen zu den Quaternionen", SpringerSpektrum, 2013