Let n be a positive integer. The factorial of n is the number n!
 obtained by multiplying all the integers among 1 and n, namely
 n! = 1·2·...·(n−1)·n.
 Observe that (n +1)! = n!·(n+1).
 Prove by induction that:
 (a) for every natural n ≥ 4,
 (b) for every natural n ≥ 2,
 n! > 2n ;
 nn >n! 

We are given two statements to prove by induction:

### (a) \( n! > 2^n \) for every natural number \( n \geq 4 \),
### (b) \( n^n > n! \) for every natural number \( n \geq 2 \).

### Proof for (a): \( n! > 2^n \) for \( n \geq 4 \)

#### **Base Case:**
Start by testing the inequality for \( n = 4 \).

- \( n! = 4! = 4 \times 3 \times 2 \times 1 = 24 \)
- \( 2^n = 2^4 = 16 \)

Clearly, \( 24 > 16 \), so the base case holds.

#### **Inductive Step:**
Now, assume that the statement is true for some \( n = k \geq 4 \), i.e., assume:

\[
k! > 2^k
\]

We want to prove that this inequality holds for \( n = k+1 \), i.e., we need to show:

\[
(k+1)! > 2^{k+1}
\]

Using the relation \( (k+1)! = k! \times (k+1) \), we get:

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

By the inductive hypothesis, we know \( k! > 2^k \), so:

\[
(k+1)! > 2^k \times (k+1)
\]

Now, we need to show that:

\[
2^k \times (k+1) > 2^{k+1}
\]

This simplifies to:

\[
(k+1) > 2
\]

This is true for \( k \geq 4 \), because for \( k \geq 4 \), \( k+1 \geq 5 \), and \( 5 > 2 \). Hence, the inequality holds for \( n = k+1 \), completing the inductive step.

Therefore, by the principle of mathematical induction, the inequality \( n! > 2^n \) holds for all \( n \geq 4 \).

---

### Proof for (b): \( n^n > n! \) for \( n \geq 2 \)

#### **Base Case:**
Test the inequality for \( n = 2 \):

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

Clearly, \( 4 > 2 \), so the base case holds.

#### **Inductive Step:**
Assume that the statement is true for some \( n = k \geq 2 \), i.e., assume:

\[
k^k > k!
\]

We want to prove that this inequality holds for \( n = k+1 \), i.e., we need to show:

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

First, express \( (k+1)! \) as:

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

By the inductive hypothesis, we know \( k^k > k! \), so:

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

Now, we need to show that:

\[
(k+1)^{k+1} > (k+1) \times k^k
\]

Dividing both sides by \( (k+1) \), this reduces to:

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

This is clearly true for \( k \geq 2 \), because \( (k+1)^k \) grows faster than \( k^k \) for all \( k \geq 2 \).

Thus, the inequality holds for \( n = k+1 \), completing the inductive step.

Therefore, by the principle of mathematical induction, the inequality \( n^n > n! \) holds for all \( n \geq 2 \).

---

### Summary:
We have proven both inequalities by mathematical induction:
1. \( n! > 2^n \) for \( n \geq 4 \),
2. \( n^n > n! \) for \( n \geq 2 \).

---

Would you like further details or have any questions? 

Here are 5 related questions to expand the concept:
1. How does the growth of \( n! \) compare to the growth of exponential functions for large \( n \)?
2. Can you apply induction to prove that \( n! \) grows faster than \( 3^n \)?
3. What is the significance of Stirling’s approximation in approximating \( n! \)?
4. Can you find other sequences that grow faster than \( n^n \)?
5. How can you generalize the inequality for other powers of \( n \) instead of \( 2^n \)?

**Tip:** Mathematical induction is a powerful tool not only for inequalities but for proving general statements about sequences and series as well!

Let n be a positive integer. The factorial of n is the number n! obtained by multiplying all the integers among 1 and n, namely n! = 1·2·...·(n−1)·n. Observe that (n +1)! = n!·(n+1). Prove by induction that: (a) for every natural n ≥ 4, n! > 2^n; (b) for every natural n ≥ 2, n^n > n!.

Explore the proof of factorial inequalities using mathematical induction. Learn how to prove that n! > 2^n for n ≥ 4 and n^n > n! for n ≥ 2 with detailed steps and induction logic. Ideal for advanced high school students (Grades 10-12).

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