Proof

The case $n=2$ was proven in the proposition about divisors of a product of two factors co-prime 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 co-prime, 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 \[a\mid \prod_{\nu=n-1}^n a_\nu,\] and finally, after the removal of $a_{n-1},$ \[a\mid a_n.\]
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Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927