Solution:

Observe that for every $i$ we have $a_{i+1} \leq a_{i}$ (every positive divisor of $a_{i}$ is less than or equal to $a_{i}$), and that $a_{i+1} = a_{i}$ if and only if $a_{i} \in \{1, 2\}$, because for $a_{i} \geq 3$ the number $a_{i} - 1$ is not a divisor of $a_{i}$.

If $a_{1} = 1$, then $a_{1}$ is a perfect square; if $a_{1} = 2$, then $a_{2} = 2$, which contradicts the hypothesis. We may therefore assume $a_{1} \geq 3$, in which case the sequence $a_{i}$ strictly decreases at each step until it reaches the value $2$. Let $k > 2$ be the first index for which $a_{k} = 2$; $a_{k-1}$ is then an odd prime, because only prime numbers have exactly two positive divisors.

Note now that if a positive integer $n$ has an odd number of divisors then it is a perfect square: if it were not, the set of its divisors could be partitioned into pairs of the form $\{a, n / a\}$ (we always have $a \neq n / a$), and hence it would have even cardinality. It follows that $a_{k-2}$ is a perfect square.