Since \(a_1|b\) and \(a_2|b\):
$$ b = a_1u $$$$ b = a_2v $$
Since \(a_1\) and \(a_2\) are coprime:
$$ a_1r + a_2s = 1 $$
We can multiply this by \(b\):
$$ (b)a_1r + (b)a_2s = b $$$$ (a_2v)a_1r + (a_1u)a_2s = b $$
Which means:
$$ a_2a_1 (vr + us) = b $$
Then \(a_2a_1|b\). We can generalize this, let's say if \(a_3|b\) and all \(a_i\) are pairwise relatively prime. If \((a_1,a_3)=1\) and \((a_2,a_3)=1\), then \((a_1a_2,a_3)=1\). If \(a_1a_2|b\), \(a_3|b\) and \((a_1a_2,a_3)=1\), then \(a_1a_2a_3|b\), by the reasoning stated above.