Proof
(related to Corollary: Divisors of a Product Of Many Factors, CoPrime to All But One Factor, Divide This Factor)
 The case $n=2$ was proven in the proposition about divisors of a product of two factors coprime to one factor.
 Let \(n > 2\). If
\[a\mid \prod_{\nu=1}^n a_\nu,\]
then it follows from \( a\perp a_\nu,\quad 1\le \nu < n,\) in particular, that \(a\) and \(a_1\) are coprime, i.e. \(\gcd(a,a_1)=1\).
 Therefore, any common divisor other of \(a\) and \(\prod_{\nu=1}^n a_\nu\) other than \(1\) must be also a common divisor of \(a\) and \(\prod_{\nu=2}^n a_\nu\), i.e. the product in which we left the first factor \(a_1\).
 Thus, it follows that
\[a\mid \prod_{\nu=2}^n a_\nu.\]
 With the same argument, since \(\gcd(a,a_2)=1\), we can remove \(a_2\) from the product and still conclude that
\[a\mid \prod_{\nu=3}^n a_\nu.\]
 The same reasoning and the same procedure can be repeated for all \(\nu=1,\ldots,n\), until we get
\[a\mid \prod_{\nu=n1}^n a_\nu,\]
and finally, after the removal of \(a_{n1},\)
\[a\mid a_n.\]
∎
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References
Bibliography
 Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927