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 $n$.