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Let's go through each part of this problem.

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**Problem Statement:** Let \( P(n) \) be the statement that \( n! < n^n \), where \( n \) is an integer greater than 1.

**Parts of the Problem:**

1. **(a) What is the statement \( P(2) \)?**

   \( P(2) \) is the statement that \( 2! < 2^2 \).

   Since \( 2! = 2 \) and \( 2^2 = 4 \), we find that \( 2 < 4 \), which is true.

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2. **(b) Show that \( P(2) \) is true, completing the basis step of a proof by mathematical induction that \( P(n) \) is true for all integers \( n \) greater than 1.**

   We have already shown in part (a) that \( P(2) \) is true because \( 2! = 2 \) and \( 2^2 = 4 \), so \( 2 < 4 \) holds. Therefore, the basis step is verified as true.

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3. **(c) What is the inductive hypothesis of a proof by mathematical induction that \( P(n) \) is true for all integers \( n \) greater than 1?**

   The inductive hypothesis is the assumption that \( P(k) \) is true for some integer \( k \geq 2 \). In other words, we assume that \( k! < k^k \) holds for a particular \( k \geq 2 \).

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4. **(d) What do you need to prove in the inductive step of a proof by mathematical induction that \( P(n) \) is true for all integers \( n \) greater than 1?**

   In the inductive step, we need to prove that if \( P(k) \) is true (i.e., if \( k! < k^k \)), then \( P(k+1) \) is also true. This means we need to show that \( (k+1)! < (k+1)^{k+1} \) under the assumption that \( k! < k^k \).

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5. **(e) Complete the inductive step of a proof by mathematical induction that \( P(n) \) is true for all integers \( n \) greater than 1.**

   **Proof of the Inductive Step:**

   - Starting from the inductive hypothesis \( k! < k^k \), we want to prove that \( (k+1)! < (k+1)^{k+1} \).

   - We can expand \( (k+1)! \) as:
     \[
     (k+1)! = (k+1) \cdot k!
     \]

   - By the inductive hypothesis, \( k! < k^k \). Substituting this into the expression, we get:
     \[
     (k+1)! = (k+1) \cdot k! < (k+1) \cdot k^k
     \]

   - Now, observe that:
     \[
     (k+1) \cdot k^k < (k+1)^{k+1}
     \]
     since \( (k+1)^{k+1} \) is a larger product by repeatedly multiplying \( k+1 \) with itself \( k+1 \) times.

   - This completes the inductive step, proving that if \( P(k) \) is true, then \( P(k+1) \) is also true.

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6. **(f) Explain why these steps show that this inequality is true whenever \( n \) is an integer greater than 1.**

   By mathematical induction, we have shown:
   - The base case \( P(2) \) is true.
   - If \( P(k) \) is true for any integer \( k \geq 2 \), then \( P(k+1) \) is also true.

   Therefore, by the principle of mathematical induction, \( P(n) \) is true for all integers \( n > 1 \).

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Would you like further details on any of these steps or have additional questions?

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### Related Questions:
1. Can you apply the same method to prove that \( n! < 2^n \) for all integers \( n > 1 \)?
2. How does the growth rate of factorial functions compare with exponential functions in general?
3. What are some common applications of mathematical induction outside of inequalities?
4. How would the proof change if we wanted to show \( n! \leq n^n \)?
5. Is it possible to use mathematical induction to prove inequalities involving other sequences, such as Fibonacci numbers?

### Tip:
Always verify the base case and clearly state the inductive hypothesis when using mathematical induction to ensure a solid proof structure.

Let P(n) be the statement that n! < n^n, where n is an integer greater than 1. Prove this statement using mathematical induction.

Explore how to use mathematical induction to prove the inequality n! < n^n for all integers n greater than 1. This solution includes the base case, inductive hypothesis, and inductive step, with clear explanations of each part. Suitable for high school students in Grades 11-12.

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