= Solution
Every multiple of $x_1\cdots x_r$ is a multiple of each $x_j$, giving one inclusion. For the reverse inclusion, a <principal ideal domain> is a <unique factorization domain>, so each <irreducible element> $x_j$ is a <prime element>.
Let $a\in\bigcap_{j=1}^r(x_j)$. The case $a=0$ is immediate. Otherwise, write $a=x_1b$. Since $x_2\mid a$ and the <prime element> $x_2$ cannot divide the non-<associate> <irreducible element> $x_1$, it divides $b$. Write $a=x_1x_2b_2$, and repeat. At stage $j$, the <prime element> $x_j$ divides none of $x_1,\ldots,x_{j-1}$, so it divides the remaining factor. Hence $x_1\cdots x_r\mid a$, proving
$$
\boxed{(x_1\cdots x_r)=(x_1)\cap\cdots\cap(x_r)}.
$$
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