Determine the integers $n, n \ge 2$, with the property that the numbers $1! , 2 ! , 3 ! , ..., (n- 1)!$ give different remainders when dividing by $n $.


1. **Identify the problem and initial conditions:**
   We need to determine the integers \( n \) (where \( n \ge 2 \)) such that the numbers \( 1!, 2!, 3!, \ldots, (n-1)! \) give different remainders when divided by \( n \).

2. **Consider the case when \( n \) is composite:**
   Let \( n = a \cdot b \) where \( a \ge b \). Since \( a \le n-1 \), we have:
   \[
   a! = 1 \cdot 2 \cdot 3 \cdots a
   \]
   Since \( a \le n-1 \), \( a! \) is a product of all integers from 1 to \( a \). Therefore, for any \( k \ge a \), \( k! \) will include \( a! \) as a factor, making \( k! \equiv 0 \pmod{n} \). This implies:
   \[
   (a+1)! \equiv (a+2)! \equiv \cdots \equiv (n-1)! \equiv 0 \pmod{n}
   \]
   For these remainders to be different, \( n \) must be such that \( n-a = 1 \), which implies \( n = 2 \).

3. **Consider the case when \( n \) is a perfect square:**
   Let \( n = x^2 \). We need to show that:
   \[
   (x+1)! \equiv (x+2)! \equiv \cdots \equiv (x^2)! \equiv 0 \pmod{n}
   \]
   Since \( x^2 \mid (x+1)! \) because:
   \[
   (x+1)! = (x-1)! \cdot x \cdot (x+1)
   \]
   and \( x \mid (x-1)! \) unless \( x \) is prime. If \( x \) is prime, we need to handle this case separately.

4. **Consider the case when \( n \) is a prime number:**
   Let \( n = p \) where \( p \) is a prime. By Wilson's Theorem, we have:
   \[
   (p-1)! \equiv -1 \pmod{p}
   \]
   and:
   \[
   (p-2)! \equiv 1 \pmod{p}
   \]
   This implies that \( 1! \equiv (p-2)! \pmod{p} \), which cannot happen since \( 1! \) and \( (p-2)! \) must be different. Therefore, we need to check \( n = 3 \), which is also a solution.

5. **Verify the solutions \( n = 2 \) and \( n = 3 \):**
   - For \( n = 2 \):
     \[
     1! \equiv 1 \pmod{2}
     \]
     The remainders are different.
   - For \( n = 3 \):
     \[
     1! \equiv 1 \pmod{3}, \quad 2! \equiv 2 \pmod{3}
     \]
     The remainders are different.

Thus, the only solutions are \( n = 2 \) and \( n = 3 \).

The final answer is \( \boxed{n \in \{2, 3\}} \).