Olimpiada Matemática del Cono Sur

In order to do this, we will show that if the sequence is alagoana, then $a_n$ cannot have any prime factor.
Consider a prime $p$. For every positive integer $n$, let $\alpha(n)$ be the exponent of $p$ in the factorization of $a_n$. We will prove that $\alpha(n) = 0$ for every $n$.
By the second condition on the sequence, we know that $\alpha(n)$ is a multiple of $n$; that is, there exists a non-negative integer $k_n$ such that $\alpha(n) = n \cdot k_n$. From the first condition, we have that $a_{n!} = a_1 \cdot a_2 \cdot \dots \cdot a_n$. This implies that $\alpha(n!) = \alpha(1) + \alpha(2) + \dots + \alpha(n)$. In particular, $\alpha(n!) = \alpha(n) + \alpha((n-1)!)$, so $\alpha(n) = \alpha(n!) - \alpha((n-1)!)$, that is, $n \cdot k_n = n! \cdot k_{n!} - (n-1)! \cdot k_{(n-1)!}$. Defining $\beta(n) = (n-1)!$, we have
$$k_n = \frac{\beta(n)}{n} (n \cdot k_{n!} - k_{\beta(n)}). \qquad (1)$$
For $n \ge 4$, we will show, recursively, that $\beta^j(n)$ divides $\alpha(n)$ for every positive integer $j$, where $\beta^j(n)$ is the function $\beta$ applied $j$ times. To this end, we will apply formula (1) recursively. We first establish some facts we will use.
Let $\gamma_m = \frac{\beta(m!)}{m!}$ be the factor appearing in formula (1) when applied to $n = m!$. Note that, for $m \ge 3$, $\gamma_m$ is an integer (since $m! - 1 > m$) and divides $k_{m!}$. In addition, $\gamma_m$ divides $\gamma_{m+1}$, since $\frac{((m+1)!-1)!}{(m+1)!} = \frac{1}{m+1} \cdot ((m+1)! - 1) ((m+1)! - 2) \cdots m! \cdot \frac{(m!-1)!}{m!}$, and $(m+1)! - (m+1) = (m+1)(m! - 1) > m!$; then, $\gamma_m$ divides $\gamma_n$ for every $n > m \ge 3$. Now, we will prove recursively that, for every positive integer $j$, we have that
$$k_n = \frac{\beta^j(n)}{n} \cdot N_j.$$
where $N_j$ is an integer that can be expressed as a sum of $2^j$ multiples of integers of the form $k_{m!}$, where the smallest subindex $m!$ that appears is $\beta^j(n)$.
For $j=1$, the statement is clear from identity (1), since $\beta(n) = (n-1)!$.
Assume the result holds for $j \ge 1$. Since every term in the expression of $N_j$ is a multiple of $k_{m!}$ for an integer $m$, and the smallest of the integers $m$ involved is $\beta^{j-1}(n) - 1$ (since, by definition, $\beta^j(n) := (\beta^{j-1}(n) - 1)!$), it follows that each term is a multiple of $\gamma_{\beta^{j-1}(n)-1} = \frac{\beta((\beta^{j-1}(n)-1)!)}{(\beta^{j-1}(n)-1)!} = \frac{\beta^{j+1}(n)}{\beta^j(n)}$. Moreover, as a consequence of identity (1) applied to $m!$, the quotient in the division of $k_{m!}$ by $\gamma_{\beta^{j-1}(n)-1}$ can be written as a sum of a multiple of $k_{(m!)!}$ and a multiple of $k_{\beta(m!)}$. Then, $N_j$ is a multiple of $\frac{\beta^{j+1}(n)}{\beta^j(n)}$, and the quotient $N_{j+1}$ can be expressed as a sum of $2^{j+1}$ terms that are multiples of integers of the form $k_{m!}$, where the smallest subindex that appears is $\beta(\beta^j(n)) = \beta^{j+1}(n)$ (note that, if $m_1 > m_2$, then $m_1! > \beta(m_1) = (m_1 - 1)! \ge m_2! > (m_2 - 1)! = \beta(m_2)$). We conclude that
$$k_n = \frac{\beta^j(n)}{n} \cdot N_j = \frac{\beta^j(n) \beta^{j+1}(n)}{n \beta^j(n)} \cdot N_{j+1} = \frac{\beta^{j+1}(n)}{n} \cdot N_{j+1},$$
as we wanted to prove.
Thus, if $n \ge 4$, for every positive integer $j$, we can write $\alpha(n) = n \cdot k_n = \beta^j(n) \cdot N_j$ for a non-negative integer $N_j$.
Finally, to conclude that $\alpha(n) = 0$ for all $n \ge 4$, we notice that $\beta^j(n) > 2^j$ for every $j$; for $j = 1$, $\beta(n) = (n - 1)! \ge 6 > 2$, since $n \ge 4$, and for $j \ge 1$, $\frac{\beta^{j+1}(n)}{\beta^j(n)} = \frac{(m - 1)!}{m} \ge 2$, since $(m - 1)! \ge 2m$ for every integer $m \ge 4$.
The remaining cases follow now easily from the identity $\alpha(1) + \alpha(2) + \alpha(3) = \alpha(3!) = \alpha(6) = 0$, which implies that $\alpha(1) = \alpha(2) = \alpha(3) = 0$.