To celebrate the $20$th LMT, the LHSMath Team bakes a cake. Each of the $n$ bakers places $20$ candles on the cake. When they count, they realize that there are $(n -1)!$ total candles on the cake. Find $n$.

[i]Proposed by Christopher Cheng[/i]
1. We start with the equation given in the problem:
   \[
   20n = (n-1)!
   \]
   We need to find the value of \( n \) that satisfies this equation.

2. We can solve this equation by trial and error, testing different values of \( n \) to see if they satisfy the equation.

3. Let's test \( n = 1 \):
   \[
   20 \cdot 1 = 20 \quad \text{and} \quad (1-1)! = 0! = 1
   \]
   Clearly, \( 20 \neq 1 \), so \( n = 1 \) is not a solution.

4. Let's test \( n = 2 \):
   \[
   20 \cdot 2 = 40 \quad \text{and} \quad (2-1)! = 1! = 1
   \]
   Clearly, \( 40 \neq 1 \), so \( n = 2 \) is not a solution.

5. Let's test \( n = 3 \):
   \[
   20 \cdot 3 = 60 \quad \text{and} \quad (3-1)! = 2! = 2
   \]
   Clearly, \( 60 \neq 2 \), so \( n = 3 \) is not a solution.

6. Let's test \( n = 4 \):
   \[
   20 \cdot 4 = 80 \quad \text{and} \quad (4-1)! = 3! = 6
   \]
   Clearly, \( 80 \neq 6 \), so \( n = 4 \) is not a solution.

7. Let's test \( n = 5 \):
   \[
   20 \cdot 5 = 100 \quad \text{and} \quad (5-1)! = 4! = 24
   \]
   Clearly, \( 100 \neq 24 \), so \( n = 5 \) is not a solution.

8. Let's test \( n = 6 \):
   \[
   20 \cdot 6 = 120 \quad \text{and} \quad (6-1)! = 5! = 120
   \]
   Here, \( 120 = 120 \), so \( n = 6 \) is a solution.

9. To ensure there are no other solutions, we note that for \( n > 6 \), the factorial function \((n-1)!\) grows much faster than the linear function \(20n\). Therefore, \( n = 6 \) is the only solution.

\(\blacksquare\)

The final answer is \( \boxed{ n = 6 } \)