For the odd $n = 2l + 1$ one can prove the statement as follows:
$$P^2 = (a_1a_2)(a_2a_3)(a_3a_4)\dots(a_{2018}a_{2019})(a_{2019}a_1) = m^{2l+1},$$
where $P = a_1a_2\dots a_{2019}$. From here it follows that $P = a_1a_2\dots a_{2019} = s^{2l+1}$. Therefore,
$$a_1 = \frac{P}{(a_2a_3)(a_4a_5)\dots(a_{2018}a_{2019})} = \frac{s^{2l+1}}{u^{2l+1}} = v_1^{2l+1},$$
and the same holds for all $a_i = v_i^{2l+1}$, $i = 2, \ldots, 2019$. From here it directly follows that
$$a_i a_j = v_i^{2l+1} v_j^{2l+1} = v^{2l+1}.$$

For even $n = 2l$ one can write
$$P^2 = (a_1a_2)(a_2a_3)(a_3a_4)\dots(a_{2018}a_{2019})(a_{2019}a_1) = m^{2l}, \text{ where } P = a_1a_2\dots a_{2019}.$$
From here it follows that $P = m^l$. Therefore,
$$a_1^2 = \frac{P^2}{(a_2a_3)(a_4a_5)\dots(a_{2018}a_{2019})} = \frac{m^{2l}}{u^{2l}} = v_1^{2l},$$
and the same holds for all $a_i^2 = v_i^{2l}$, $i = 2, \ldots, 2019$. From here it directly follows that
$$r^{2l} = (a_1a_2)(a_2a_3)(a_3a_4)\dots(a_{k-1}a_k) = a_1a_2^2a_3^2\dots a_{k-1}a_k = a_1a_k w^{2l} \Rightarrow a_1a_k = \frac{r^{2l}}{w^{2l}} = z^{2l}.$$