1. **State the problem:** We want to prove that for all positive integers $n$, the product $n(n+1)(n+2)$ is divisible by 6. 2. **Understand divisibility by 6:** A number is divisible by 6 if and only if it is divisible by both 2 and 3. 3. **Check divisibility by 2:** Among three consecutive integers $n$, $n+1$, and $n+2$, at least one must be even, so the product includes a factor of 2. 4. **Check divisibility by 3:** Among three consecutive integers, exactly one is divisible by 3 because the remainders when dividing by 3 cycle through 0, 1, 2. Therefore, the product includes a factor of 3. 5. **Combine results:** Since the product has at least one factor of 2 and at least one factor of 3, it must be divisible by $2 \times 3 = 6$. 6. **Conclusion:** Hence, $n(n+1)(n+2)$ is divisible by 6 for all positive integers $n$. \boxed{\text{The product } n(n+1)(n+2) \text{ is divisible by } 6 \text{ for all } n \in \mathbb{Z}^+.}