If $\gcd(a_i, a_j)=1$ for $i\ne j$ then $\gcd(a_2 ... a_n, ..., a_1 ... a_{n-1})=1$?

Question

I try to prove this statement for $i,j \in \{1,...,n\}$.

I notice that for each term for instance $a_1$ divides $(n-1)$ terms and it is the same for each $a_k$ but I think it is not enough to conclude.

I also try to use Bézout's theorem for each $(a_i,a_j)$ and multiply these equations but no success.

Thanks in advance.

Answer 1

HINT:

Assume there exists a prime that divides all the products of $n-1$ elements. Now, since it divides $a_2\ldots a_n$ it must divide one of the numbers $a_2$, $\ldots$ , $a_n$. Assume that number is $a_n$. Now that prime must divide the product where $a_n$ does not appear, so one of the numbers $a_1$, $\ldots$, $a_{n-1}$, contradiction.

Obs: You can also multiply all the equations from Bezout if you look carefully at the terms obtained.