Proof
(related to Proposition: Existence of Prime Divisors)
 Trivially, every prime divides the integer $0.$ Also, because with every divisor \(d\mid n\) it is also true that \(d\mid 1\), it is sufficient to show the existence of prime divisors for natural numbers \(n > 1\), instead of integers \(n\neq\pm 1\).
 Let \(D(n):=\{d\in\mathbb N:b>1,~d\mid n\}\) be the set of all divisors of \(n\).
 \(D(n)\) is not empty, since \(n\in D(n)\).
 Therefore, according to the wellordering principle, \(D(n)\) must have a smallest element, which we denote by \(p\).
 By construction, \(p > 1\).
 We will now show that \(p\) is a prime number.
 For if it was composite, then it would have a nontrivial divisor \(q\mid p\), i.e. a divisor with \(1 < q < p\).
 Because \(q\mid p\) and \(p\mid n\), it would follow from the divisibility laws that \(q\mid n\) and so \(q\in D(n)\).
 This is a contradiction to the minimality of \(p\) in \(D(n)\).
 Therefore, \(p\) must be prime.
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References
Bibliography
 Kramer Jürg, von Pippich, AnnaMaria: "Von den natürlichen Zahlen zu den Quaternionen", SpringerSpektrum, 2013