$\gcd(a_1,...,a_n)=\gcd(\gcd(a_1,...,a_{n-1}),a_n)$

Question

If I define $\gcd(a_1,...,a_n)$ be the greatest positive common divisor or $a_1,...,a_n$.

How can I show $\gcd(a_1,...,a_n)=\gcd(\gcd(a_1,...,a_{n-1}),a_n)$?

The equality is trivial if we have $\forall $ common divisor $c$ of $a_1,...,a_n$, $c\mid \gcd(a_1,...,a_n)$. But I think this is a consequence of the above proposition by extending the result from the case $n=2$.

It is easy to show $\gcd(\gcd(a_1,...,a_{n-1}),a_n)\le \gcd(a_1,...,a_n),$ but I got stuck showing the another side. Any suggestion will be appreciated.

Answer 1

Hint: Prove more generally that $D(a_1,\dots,a_n)=D(a_1,\dots,a_{n-1}) \cap D(a_n)$, where $D(S)$ is the set of all positive common divisors of all elements in a set $S$. Then, by definition, $\gcd(S)=\max D(S)$. Furthermore, $D(S)=D(\gcd(S))$.