Find all prime numbers $p$, for which the number $p + 1$ is equal to the product of all the prime numbers which are smaller than $p$.
1. **Understanding the Problem:**
   We need to find all prime numbers \( p \) such that \( p + 1 \) is equal to the product of all prime numbers smaller than \( p \).

2. **Using Bonse's Inequality:**
   Bonse's inequality states that for \( n \geq 5 \), the product of the first \( n-1 \) primes is greater than the square of the \( n \)-th prime:
   \[
   p_1 p_2 \cdots p_{n-1} > p_n^2
   \]
   Here, \( p_n \) denotes the \( n \)-th prime number.

3. **Applying Bonse's Inequality:**
   Suppose \( p \) is the \( n \)-th prime number. Then, we have:
   \[
   p + 1 = p_1 p_2 \cdots p_{n-1}
   \]
   According to Bonse's inequality, for \( n \geq 5 \):
   \[
   p_1 p_2 \cdots p_{n-1} > p_n^2
   \]
   This implies:
   \[
   p + 1 > p^2
   \]
   which is impossible because \( p + 1 \leq p^2 \) for \( p \geq 5 \).

4. **Checking Smaller Primes:**
   Since \( n \geq 5 \) leads to a contradiction, we need to check the cases for \( n \leq 4 \):
   - For \( n = 1 \): \( p = 2 \)
     \[
     p + 1 = 2 + 1 = 3 \quad \text{and} \quad \text{product of primes less than } 2 = 1 \quad (\text{no primes less than } 2)
     \]
     This does not satisfy the condition.
   - For \( n = 2 \): \( p = 3 \)
     \[
     p + 1 = 3 + 1 = 4 \quad \text{and} \quad \text{product of primes less than } 3 = 2
     \]
     This does not satisfy the condition.
   - For \( n = 3 \): \( p = 5 \)
     \[
     p + 1 = 5 + 1 = 6 \quad \text{and} \quad \text{product of primes less than } 5 = 2 \times 3 = 6
     \]
     This satisfies the condition.
   - For \( n = 4 \): \( p = 7 \)
     \[
     p + 1 = 7 + 1 = 8 \quad \text{and} \quad \text{product of primes less than } 7 = 2 \times 3 \times 5 = 30
     \]
     This does not satisfy the condition.

5. **Conclusion:**
   The only prime number \( p \) for which \( p + 1 \) is equal to the product of all prime numbers smaller than \( p \) is \( p = 5 \).

The final answer is \( \boxed{ p = 5 } \).