Let \((a_1, a_2, \ldots) = c\) and \((ma_1, ma_2, ...) = d\), so:
\[ \begin{gather} c|a_1 \wedge c|a_2 \wedge \cdots \\ \implies mc|ma_1 \wedge mc|ma_2 \wedge \cdots \end{gather} \]
Since \(mc\) divides \(ma_1, ma_2, \ldots\), and since \(d\) is the greatest common divisor of \(ma_1, ma_2, \ldots\), then \(mc\) has to be some multiple of \(d\):
$$ d = mcx $$
for some integer \(x\), so:
\[ \begin{gather} d|ma_1 \wedge d|ma_2 \wedge \cdots \implies \\ mcx|ma_1 \wedge mcx|ma_2 \wedge \cdots \\ \therefore cx|a_1 \wedge cx|a_2 \wedge \cdots \end{gather} \]
Since \(c\) is the greatest common divisor of \(a_1, a_2, \ldots\), then \(x=1\), so \(d = mc\).