$$4d_{i+1}^2 - 4d_{i+2} \cdot d_i \ge 0 \Leftrightarrow \frac{d_{i+1}}{d_i} \ge \frac{d_{i+2}}{d_{i+1}} \quad (*)$$
for any $i \in \{1, 2, \dots, k-2\}$.
As $d_2$ is the smallest proper divisor of $n$, it means that the number $\frac{n}{d_2}$ is the greatest proper divisor of $n$, so $\frac{n}{d_2} = d_{k-1}$. We have:
$$\frac{n}{d_2} = d_{k-1} = \prod_{i=1}^{k-2} \frac{d_{i+1}}{d_i} \ge \prod_{i=1}^{k-2} \frac{d_{i+2}}{d_{i+1}} = \frac{n}{d_2}.$$
It follows that all inequalities (*) turn into equalities.
Thus, $d_{i+1}^2 = d_{i+2} \cdot d_i$ for any $i \in \{1, 2, \dots, k-2\}$. It follows that the numbers $1 = d_1, d_2, d_3, \dots, d_k = n$ (in this order) are consecutive terms of a geometric progression of ratio $d_2$, so $n = d_2^{k-1}$.
The lowest proper divisor of the composite number $n$ is a prime number $p$, so $d_2 = p$, and $n = p^{k-1}$.