Macedonian Mathematical Olympiad

We'll write $a \leftrightarrow b$ if there exists such a sequence. It is easy to see that $\leftrightarrow$ is an equivalence relation. Notice that $n \leftrightarrow 2n$ for every integer $n$, $n \ge 3$ because in that case the desired finite sequence is
$$n_1 = n,\ n_2 = n(n-1),\ n_3 = n(n-1)(n-2),\ n_4 = n(n-2),\ n_5 = 2n$$
For every $n$, $n \ge 4$ we have $n' = (n-1)(n-2) \ge 3$ and we obtain $n' \leftrightarrow 2n'$. For $n \ge 4$:
$$\begin{aligned} n_1 &= n,\ n_2 = n(n-1),\ n_3 = n(n-1)(n-2),\ n_4 = n(n-1)(n-2)(n-3),\\ n_5 &= 2(n-1)(n-2) = 2n' \end{aligned}$$
i.e. $n \leftrightarrow 2n'$ and because $n_1' = n' = (n-1)(n-2)$, $n_2' = n-1$ we obtain $n' \leftrightarrow n-1$. From the previous discussion we get that $n \leftrightarrow 2n'$, $2n' \leftrightarrow n'$, $n' \leftrightarrow n-1$ and from the transitivity of $\leftrightarrow$ we obtain $n \leftrightarrow n-1$ for every integer $n$, $n \ge 4$. Because $\leftrightarrow$ is symmetric $n \leftrightarrow n+1$ for every integer $n$, $n \ge 3$. Hence from the transitivity of $\leftrightarrow$ we obtain the desired result.