Prove that there exists a positive integer N such that for all n ≥ N , we have2023n ≤ n!. You may use, without proof, the fact that 20236000 ≤ 6000

Question

Prove that there exists a positive integer N such that for all n ≥ N , we have2023n ≤ n!. You may use, without proof, the fact that 20236000 ≤ 6000
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Solution 1

To prove this, we can use the concept of mathematical induction.

Step 1: Base Case Let's start with n = 2024. We have 20232024 ≤ 2024!. This is true because 2024! is a product of 2024 numbers, each of which is at least 1, and one of which is 2024. So, 2024! is at least 2024111...*1 (2023 times) Knowee AI StudyGPT is a powerful AI-powered study tool designed to help you to solve study prob

Knowee AI StudyGPT is a powerful AI-powered study tool designed to help you to solve study problem.
Knowee AI StudyGPT is a powerful AI-powered study tool designed to help you to solve study problem.
Knowee AI StudyGPT is a powerful AI-powered study tool designed to help you to solve study problem.
Knowee AI StudyGPT is a powerful AI-powered study tool designed to help you to solv

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