Find all non-negative integers $n$, $a$, and $b$ satisfying
\[2^a + 5^b + 1 = n!.\]
1. **Initial Consideration:**
   We start by noting that \( n! \geq 3 \) implies \( n > 2 \). This is because the smallest factorial greater than or equal to 3 is \( 3! = 6 \).

2. **Case \( n \geq 4 \):**
   - **Modulo 4 Analysis:**
     For \( n \geq 4 \), \( n! \) is divisible by 4. Therefore, \( n! \equiv 0 \pmod{4} \).
     \[
     2^a + 5^b + 1 \equiv 0 \pmod{4}
     \]
     Since \( 5 \equiv 1 \pmod{4} \), we have:
     \[
     2^a + 1 + 1 \equiv 0 \pmod{4} \implies 2^a + 2 \equiv 0 \pmod{4} \implies 2^a \equiv 2 \pmod{4}
     \]
     This implies \( a = 1 \) because \( 2^1 \equiv 2 \pmod{4} \) and for \( a \geq 2 \), \( 2^a \equiv 0 \pmod{4} \).

   - **Modulo 5 Analysis:**
     For \( n \geq 5 \), \( n! \) is divisible by 5. Therefore, \( n! \equiv 0 \pmod{5} \).
     \[
     2^1 + 5^b + 1 \equiv 0 \pmod{5}
     \]
     Since \( 2^1 = 2 \), we have:
     \[
     2 + 5^b + 1 \equiv 0 \pmod{5} \implies 3 + 5^b \equiv 0 \pmod{5} \implies 5^b \equiv -3 \equiv 2 \pmod{5}
     \]
     However, \( 5^b \equiv 0 \pmod{5} \) for \( b \geq 1 \), which is a contradiction. Therefore, \( n \geq 5 \) is not possible.

3. **Case \( n = 3 \):**
   For \( n = 3 \), we have:
   \[
   2^a + 5^b + 1 = 3!
   \]
   \[
   2^a + 5^b + 1 = 6 \implies 2^a + 5^b = 5
   \]
   - **Checking Possible Values:**
     - If \( a = 0 \), \( 2^0 = 1 \):
       \[
       1 + 5^b = 5 \implies 5^b = 4 \quad \text{(no integer solution for } b\text{)}
       \]
     - If \( a = 1 \), \( 2^1 = 2 \):
       \[
       2 + 5^b = 5 \implies 5^b = 3 \quad \text{(no integer solution for } b\text{)}
       \]
     - If \( a = 2 \), \( 2^2 = 4 \):
       \[
       4 + 5^b = 5 \implies 5^b = 1 \implies b = 0
       \]
       This gives the solution \( (a, b, n) = (2, 0, 3) \).

4. **Case \( n = 2 \):**
   For \( n = 2 \), we have:
   \[
   2^a + 5^b + 1 = 2!
   \]
   \[
   2^a + 5^b + 1 = 2 \implies 2^a + 5^b = 1
   \]
   - **Checking Possible Values:**
     - If \( a = 0 \), \( 2^0 = 1 \):
       \[
       1 + 5^b = 1 \implies 5^b = 0 \quad \text{(no integer solution for } b\text{)}
       \]
     - If \( a \geq 1 \), \( 2^a \geq 2 \):
       \[
       2^a + 5^b \geq 2 \quad \text{(no solution)}
       \]

5. **Case \( n = 1 \):**
   For \( n = 1 \), we have:
   \[
   2^a + 5^b + 1 = 1!
   \]
   \[
   2^a + 5^b + 1 = 1 \implies 2^a + 5^b = 0
   \]
   - **Checking Possible Values:**
     - If \( a \geq 0 \) and \( b \geq 0 \):
       \[
       2^a + 5^b \geq 1 \quad \text{(no solution)}
       \]

Conclusion:
The only solution is \( (a, b, n) = (2, 0, 3) \).

The final answer is \( \boxed{ (a, b, n) = (2, 0, 3) } \).