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To prove that \( n! < n^n \) for all \( n > 1 \) using mathematical induction, we follow these steps:

### Step 1: Base Case
For \( n = 2 \):

- \( 2! = 2 \)
- \( 2^2 = 4 \)

Clearly, \( 2! = 2 < 4 = 2^2 \), so the inequality holds for \( n = 2 \).

### Step 2: Inductive Hypothesis
Assume that the inequality \( n! < n^n \) holds for some \( n = k \), i.e.,

\[
k! < k^k
\]

### Step 3: Inductive Step
We need to prove that the inequality holds for \( n = k + 1 \), i.e.,

\[
(k+1)! < (k+1)^{k+1}
\]

We know that:

\[
(k+1)! = (k+1) \cdot k!
\]

From the inductive hypothesis, we have \( k! < k^k \), so:

\[
(k+1)! = (k+1) \cdot k! < (k+1) \cdot k^k
\]

Now, we need to show that:

\[
(k+1) \cdot k^k < (k+1)^{k+1}
\]

This simplifies to checking whether:

\[
k^k < (k+1)^k
\]

Since \( k + 1 > k \), we know that \( (k+1)^k > k^k \) for all \( k > 1 \). Thus, the inequality holds, and therefore:

\[
(k+1)! < (k+1)^{k+1}
\]

### Conclusion
By the principle of mathematical induction, \( n! < n^n \) holds for all \( n > 1 \).

---

Would you like more details or have any specific questions on this? Here are 5 related questions to consider:

1. How can we further verify the base case for different values of \( n \)?
2. Why is the inequality \( (k+1)^k > k^k \) always true for \( k > 1 \)?
3. What are other methods to prove inequalities like \( n! < n^n \)?
4. Can we extend this inequality to \( n \leq 1 \), and why or why not?
5. How does the behavior of factorials compare to exponential functions as \( n \) grows larger?

**Tip:** When working with factorials and exponentials, recall that factorial growth is slower than exponential growth for large \( n \).

Show that n! < n^n for all n > 1 using induction.

Learn how to prove the inequality n! < n^n for all n > 1 using the principle of mathematical induction. The solution covers key concepts of factorials, exponentiation, and inequalities. Suitable for high school students, especially those in Grades 10-12.

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