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Factorial notation is a crucial concept in mathematics, and it is represented by an exclamation mark \(!\). The factorial of a positive integer \(n\), written as \(n!\), is the product of all positive integers less than or equal to \(n\). So, \(n! = n \times (n-1) \times (n-2) \times \ldots \times 2 \times 1\). For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).In our problem, we are using \((n+1)!\), which is the factorial of \((n+1)\). This product includes all integers from 1 to \((n+1)\), inclusive. This notation simplifies our expression, making it easier to manipulate in the proofs.

Proof techniques are strategies used to establish the truth of mathematical statements. In our exercise, the goal is to prove that there are \(n\) consecutive composite integers for any positive integer \(n\).We based our proof on the sequence starting at \((n+1)! + 2\). By showing that every number in this sequence is composite, we use a technique called direct proof.One key element of the proof is identifying a structure or pattern that guarantees the composite nature of the numbers. Here, we identified a sequence of \((n+1)! + k\) for \(k\) ranging from 2 to \(n+1\). We demonstrated that each term in this sequence is composite by leveraging the properties of factorial and divisibility.

Understanding the properties of integers is fundamental when working with proofs. Integers include whole numbers and their negatives, such as -3, 0, 1, and 4. Important properties include divisibility, even and odd classifications, and prime versus composite classification.In this proof, we rely on the compositeness of the terms in our sequence. A composite number is an integer greater than 1 that has at least one positive divisor other than 1 and itself. By structuring our sequence appropriately, we can demonstrate that each term is divisible by an integer \(k \geq 2\), confirming their compositeness.We also make use of factorial properties. Since \((n+1)!\) contains all numbers from 1 to \(n+1\), adding a smaller integer \(k\) to it will still maintain divisibility by \(k\). These properties of integers underpin our direct proof methodology.

Divisibility rules help determine if one integer divides another without leaving a remainder. For example, an integer is divisible by 2 if its last digit is even.In our problem, we use the divisibility principle to show that \((n+1)! + k\) is divisible by \(k\). Since \((n+1)!\) is a multiple of all numbers from 1 to \(n+1\), adding \(k\) (which also ranges from 2 to \(n+1\)) ensures \((n+1)! + k\) remains divisible by \(k\).For example, with \(n = 4\), we have \((4+1)! = 5! = 120\). If we take \(k = 3\), then \(120 + 3 = 123\). Since 123 is divisible by 3, it is composite. This principle applies for all \(k\) in our range, confirming that the entire sequence consists of composite numbers. Understanding these rules makes it easier to identify and verify the compositeness of numbers in our proof.