A sequence $a_1, a_2,\dots, a_n$ of positive integers is [i]alagoana[/i], if for every $n$ positive integer, one have these two conditions
I- $a_{n!} = a_1\cdot a_2\cdot a_3\cdots a_n$
II- The number $a_n$ is the $n$-power of a positive integer.
Find all the sequence(s) [i]alagoana[/i].
1. **Define the sequence and conditions:**
   We are given a sequence \( \{a_n\} \) of positive integers that satisfies two conditions:
   - \( a_{n!} = a_1 \cdot a_2 \cdot a_3 \cdots a_n \)
   - \( a_n \) is the \( n \)-th power of a positive integer.

2. **Transform the sequence:**
   Let \( a_n = b_n^n \) where \( b_n \) is a positive integer. This transformation is based on the second condition that \( a_n \) is the \( n \)-th power of a positive integer.

3. **Substitute into the first condition:**
   Using the first condition, we have:
   \[
   a_{n!} = a_1 \cdot a_2 \cdot a_3 \cdots a_n
   \]
   Substituting \( a_n = b_n^n \), we get:
   \[
   a_{n!} = b_{n!}^{n!} = b_1^1 \cdot b_2^2 \cdot b_3^3 \cdots b_n^n
   \]

4. **Equate the exponents:**
   Since \( a_{n!} = b_{n!}^{n!} \), we need:
   \[
   b_{n!}^{n!} = b_1^1 \cdot b_2^2 \cdot b_3^3 \cdots b_n^n
   \]
   Taking the \( n! \)-th root on both sides, we get:
   \[
   b_{n!} = \left( b_1^1 \cdot b_2^2 \cdot b_3^3 \cdots b_n^n \right)^{1/n!}
   \]

5. **Simplify the expression:**
   For the equality to hold for all \( n \), \( b_{n!} \) must be a constant value. Let \( b_{n!} = c \) for some constant \( c \). Then:
   \[
   c = \left( b_1^1 \cdot b_2^2 \cdot b_3^3 \cdots b_n^n \right)^{1/n!}
   \]

6. **Determine the constant \( c \):**
   Since \( b_n \) must be a positive integer and the product of powers must be consistent, the only possible value for \( b_n \) that satisfies this for all \( n \) is \( b_n = 1 \). Thus:
   \[
   a_n = 1^n = 1
   \]

7. **Verify the solution:**
   If \( a_n = 1 \) for all \( n \), then:
   \[
   a_{n!} = 1 = a_1 \cdot a_2 \cdot a_3 \cdots a_n = 1 \cdot 1 \cdot 1 \cdots 1
   \]
   This satisfies both conditions given in the problem.

The final answer is \( \boxed{\{1\}_{n \in \mathbb{Z}_{>0}}} \).