Ukrainian National Mathematical Olympiad

Let $a_1 = p$ is prime. Then it is clear that
$$\begin{array}{ccc} a_2 \vdash p, & a_3 \vdash a_2 a_1 \vdash p^2, & a_4 \vdash a_3 a_2 a_1 \vdash p^4, \\ a_5 \vdash a_4 a_3 a_2 a_1 \vdash p^8, & a_6 \vdash a_5 a_4 a_3 a_2 a_1 \vdash p^{16}. \end{array}$$
Assume the length of the sequence is greater than 6, thus $a_7 \vdash a_6 a_5 a_4 a_3 a_2 a_1 \vdash p^{32}$, there is a prime number that divides $a_7$ and it is included with degree not less than 32. By conditions on $a_n$ that is not possible. Therefore, the maximum length is $n=6$. The only prime number that is included with degree 16 in $a_6 = 2^2 \cdot 3^3 \cdot 5^5 \cdot 7^7 \cdot 11^{11} \cdot 13^{13} \cdot 17^{17}$ is $p=17$. This is the maximum value of $a_1$.

It suffices to show now, that such sequence exists. Take
$$a_1 = 17,\ a_2 = 17,\ a_3 = 17^2,\ a_4 = 17^4,\ a_5 = 17^8,\ a_6 = 2^2 \cdot 3^3 \cdot 5^5 \cdot 7^7 \cdot 11^{11} \cdot 13^{13} \cdot 17^{17}.$$