Exercise 1.2.37 ($(a_1,a_2,\ldots,a_n) = ((a_1,a_2,\ldots,a_{n-1}),a_n)$)

Proof. Let $d=a_1\wedge a_2\wedge \cdots \wedge a_n$, and $\delta =(a_1\wedge a_2\wedge \cdots \wedge a_{n-1})\wedge a_n$.

Then $d\mid a_1,\dots ,d\mid a_{n-1}$, thus $d\mid a_1\wedge a_2\wedge \cdots \wedge a_{n-1}$. Moreover, $d\mid a_n$, therefore $d\mid (a_1\wedge a_2\wedge \cdots \wedge a_{n-1})\wedge a_n=\delta$.

On the other hand, $\delta \mid a_1\wedge a_2\wedge \cdots \wedge a_{n-1}$, thus $\delta \mid a_1,\delta \mid a_2,\dots ,\delta \mid a_{n-1}$. Moreover, $\delta \mid a_n$, therefore $\delta \mid a_1\wedge a_2\wedge \cdots \wedge a_n=d$.

From $d\mid \delta$ and $\delta \mid d$, where $d\ge 0,\delta \ge 0$, we deduce $d=\delta$, so

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