Answer: The smallest common difference is $n!$.

Solution 1. The condition in the problem is equivalent to
$$D=\left(b_{1}-a_{1}\right) a_{2} a_{3} \cdots a_{n}=b_{1}\left(b_{2}-a_{2}\right) a_{3} a_{4} \cdots a_{n}=\cdots=b_{1} b_{2} \cdots b_{n-1}\left(b_{n}-a_{n}\right),$$
where $D$ is the common difference. Since the progression is strictly increasing, $D>0$, hence $b_{i}>a_{i}$ for every $1 \leqslant i \leqslant n$. Individually, these equalities simplify to
$$\begin{equation*} \left(b_{i}-a_{i}\right) a_{i+1}=b_{i}\left(b_{i+1}-a_{i+1}\right) \text{ for every } 1 \leqslant i \leqslant n-1 \tag{1} \end{equation*}$$
If $g_{i}:=\operatorname{gcd}\left(a_{i}, b_{i}\right)>1$ for some $1 \leqslant i \leqslant n$, then we can replace $a_{i}$ with $\frac{a_{i}}{g_{i}}$ and $b_{i}$ with $\frac{b_{i}}{g_{i}}$ to get a smaller common difference. Hence we may assume $\operatorname{gcd}\left(a_{i}, b_{i}\right)=1$ for every $1 \leqslant i \leqslant n$.
Then, we have $\operatorname{gcd}\left(b_{i}-a_{i}, b_{i}\right)=\operatorname{gcd}\left(a_{i}, b_{i}\right)=1$ and $\operatorname{gcd}\left(a_{i+1}, b_{i+1}-a_{i+1}\right)=\operatorname{gcd}\left(a_{i+1}, b_{i+1}\right)=1$ for every $1 \leqslant i \leqslant n-1$. The equality (1) implies $a_{i+1}=b_{i}$ and $b_{i}-a_{i}=b_{i+1}-a_{i+1}$. Thus,
$$a_{1}, \quad b_{1}=a_{2}, \quad b_{2}=a_{3}, \quad \ldots, \quad b_{n-1}=a_{n}, \quad b_{n}$$
is an arithmetic progression with positive common difference. Since $a_{1} \geqslant 1$, we have $a_{i} \geqslant i$ for every $1 \leqslant i \leqslant n$, so
$$D=\left(b_{1}-a_{1}\right) a_{2} a_{3} \cdots a_{n} \geqslant 1 \cdot 2 \cdot 3 \cdots n=n!$$
Equality is achieved when $b_{i}-a_{i}=1$ for $1 \leqslant i \leqslant n$ and $a_{1}=1$, i.e. $a_{i}=i$ and $b_{i}=i+1$ for every $1 \leqslant i \leqslant n$. Indeed, it is straightforward to check that these integers produce an arithmetic progression with common difference $n!$.

Solution 2 (Variant of Solution 1). Similarly to Solution 1, we may assume $\operatorname{gcd}\left(a_{i}, b_{i}\right)=1$ for every $1 \leqslant i \leqslant n$.
Denote by $p_{1}, p_{2}, \ldots, p_{n+1}$ the sequence obtained as the product in the problem statement. Then we have $\frac{p_{i+1}}{p_{i}}=\frac{b_{i}}{a_{i}}>1$, so $b_{i}>a_{i}$. Since $p_{1}, p_{2}, \ldots, p_{n+1}$ is an arithmetic progression, we have $p_{i+2}=2 p_{i+1}-p_{i}$ hence
$$2-\frac{a_{i}}{b_{i}}=\frac{2 b_{i}-a_{i}}{b_{i}}=\frac{2 p_{i+1}-p_{i}}{p_{i+1}}=\frac{p_{i+2}}{p_{i+1}}=\frac{b_{i+1}}{a_{i+1}} .$$
But since the fractions on the left-hand side and the right-hand side are both irreducible, we conclude that $b_{i}=a_{i+1}$, so $2-\frac{a_{i}}{a_{i+1}}=\frac{a_{i+2}}{a_{i+1}}$. Then we have $a_{i}+a_{i+2}=2 a_{i+1}$, which means that $a_{1}, a_{2}, \ldots, a_{n}$ is an arithmetic progression with positive common difference.
We conclude as in Solution 1.

Solution 3. (The following solution is purely algebraic: it does not involve considerations on greatest common divisors.)
We retake Solution 1 from (1). Then we have
$$\frac{a_{i+1}}{b_{i+1}-a_{i+1}}=\frac{b_{i}}{b_{i}-a_{i}}=1+\frac{a_{i}}{b_{i}-a_{i}} .$$
So, for $1 \leqslant i \leqslant n$,
$$\frac{a_{i}}{b_{i}-a_{i}}=\frac{a_{1}}{b_{1}-a_{1}}+(i-1)$$
Then
$$a_{i} \geqslant \frac{a_{i}}{b_{i}-a_{i}}=\frac{a_{1}}{b_{1}-a_{1}}+(i-1)>i-1$$
since $b_{i}-a_{i} \geqslant 1$ and $b_{1}-a_{1}>0$. As $a_{i}$ is an integer, we have $a_{i} \geqslant i$.
We again conclude as in Solution 1.