# 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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### References

#### Bibliography

1. Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927