Final Answer:
The statement is true because the product of any k successive integers includes all the factors of k!, ensuring its divisibility by k!.
Step-by-step explanation:
To understand why the statement is true, let's consider a sequence of k successive integers, starting from n and going up to n + k - 1. The product of these k integers can be expressed as n × (n + 1) × (n + 2) × ... × (n + k - 1). We want to show that this product is divisible by k!.
Now, k! (k factorial) is the product of all positive integers up to k. Therefore, k! = 1 × 2 × 3 × ... × k. If we look at our product of k successive integers, we notice that each term in the product is one of the factors in k!. For example, if n = 1, then our product is 1 × 2 × 3 × ... × k, which is exactly k!. Similarly, for any other starting value of n, the product includes the same factors as k!.
In conclusion, since the product of any k successive integers includes all the factors of k!, it is divisible by k!. This establishes the truth of the statement. Understanding these concepts is fundamental in number theory and combinatorics, providing insights into the relationships between factorials and products of successive integers.