The smallest possible common difference is $n!$. Note that the common difference is
$$D = (b_1 - a_1)a_2a_3 \cdots a_n = b_1(b_2 - a_1)a_3a_4 \cdots = \cdots = b_1b_2 \cdots b_{n-1}(b_n - a_n).$$
Moreover, since the sequence is strictly increasing, $D > 0$, and hence $b_i > a_i$ holds for all $i$. Thus the above equation is equivalent to
$$(b_i - a_i)a_{i+1} = b_i(b_{i+1} - a_{i+1}) \text{ holding for all } i.$$
Furthermore, since if $k_i = \text{gcd}(a_i, b_i)$, then replacing $a_i$ and $b_i$ by $a_i/k_i$ and $b_i/k_i$ still satisfies the conditions of the problem, we may assume without loss of generality that $\text{gcd}(a_i, b_i) = 1$ holds for all $i$. In this case, the above equation forces us to have $b_i - a_i = b_{i+1} - a_{i+1}$ and $a_{i+1} = b_i$. Thus,
$$a_1, a_2 = b_1, a_3 = b_2, \cdots, a_n = b_{n-1}, b_n$$
must be a strictly increasing arithmetic progression. This means that, if the common difference of this sequence is $d \ge 1$, then
$$D = d \prod_{i=1}^{n-1} (a_1 + id) \ge 1 \times \prod_{i=1}^{n-1} (1 + i) = n!.$$
Equality holds when $a_i = i$.