I have the following question with me:
"Let $a, b$ be integers greater than 2. Prove that there exists a positive integer $k$ and a finite sequence $n_1, n_2, \ldots, n_k$ of positive integers such that $n_1 = a$, $n_k = b$, and $n_in_{i+1}$ is divisible by $n_i + n_{i+1}$ for each $i$ ($1 \le i \le k$)."
The solution for which is given at
https://artofproblemsolving.com/wiki/index.php?title=2002_USAMO_Problems/Problem_5
In the above link, there is this step where we have
$a \leftrightarrow \prod_{i=b}^{a}i \leftrightarrow (b-1)^{\lambda} \prod_{i=b}^{a}i \leftrightarrow (b-1)^{\lambda} \prod_{i=b}^{a}i \cdot \prod_{i=a+1}^{(b-1)^{\lambda}-1} i = X$
The first part is clear since he considers a to be number and uses the property he proves above considering a itself to be a divisor. How does the second part of this step come up? How did he conclude that there does exist such a sequence between $ (b-1)^{\lambda} \prod_{i=b}^{a}i$ and $\prod_{i=b}^{a}i$?
Let $n=\prod_{i=b}^{a}i$ and $d=b$ and from the solution, we know that $n \leftrightarrow n(d-1)^{\lambda}$ for any $\lambda\in \mathbf{N}$ so $\prod_{i=b}^{a}i \leftrightarrow (b-1)^{\lambda} \prod_{i=b}^{a}i$.