\(a_{n+1}=n+1+\dfrac{(n-1)!}{a_1a_2\cdots a_n} \;(n\geq 1)\).
The following is a part of the process of computing the general term \(a_n\).
The following is true for all positive integers \(n\).
\(a_1a_2\cdots a_n a_{n+1}\)
\(= a_1a_2\cdots a_n\times(n+1)+(n-1)!\)
Let \(b_n=\dfrac{a_1a_2\cdots a_n}{n!}\).
Then, \(b_1=1\) and
\(b_{n+1}=b_n + \fbox{\(\;(\alpha)\;\)}\).
The general term of the sequence \(\{b_n\}\) is
\(b_n=\fbox{\(\;(\beta)\;\)}\),
therefore \(\dfrac{a_1a_2\cdots a_n}{n!}=\fbox{\(\;(\beta)\;\)}\).
\(\vdots\)
Therefore \(a_1=1\), and
\(a_n=\dfrac{(n-1)(2n-1)}{2n-3} \: (n\geq 2)\).
Let \(f(n)\) be the correct expression for \((\alpha)\),
and \(g(n)\) be the correct expression for \((\beta)\).
What is the value of \(f(13)\times g(7)\)?