For natural number n , `2^n (n-1)!lt n^n` , if

Options

A. `n lt 2`
B. `n gt 2`
C. `n ge 2`
D. `n gt 3`

Answer

To solve the inequality \( 2^n (n-1)! < n^n \) for natural numbers \( n \), we will test the values of \( n \) starting from 1 and moving upwards until we find a pattern or determine the range of \( n \) for which the inequality holds true.

### Step-by-Step Solution:

1. **Test \( n = 1 \)**:
\[ 2^1 (1-1)! < 1^1 \]
This simplifies to:
\[ 2 \cdot 0! < 1 \quad \Rightarrow \quad 2 \cdot 1 < 1 \quad \Rightarrow \quad 2 < 1 \quad \text{(False)} \]
Therefore, the inequality does not hold for \( n = 1 \).

2. **Test \( n = 2 \)**:
\[ 2^2 (2-1)! < 2^2 \]
This simplifies to:
\[ 4 \cdot 1! < 4 \quad \Rightarrow \quad 4 \cdot 1 < 4 \quad \Rightarrow \quad 4 < 4 \quad \text{(False)} \]
Therefore, the inequality does not hold for \( n = 2 \).

3. **Test \( n = 3 \)**:
\[ 2^3 (3-1)! < 3^3 \]
This simplifies to:
\[ 8 \cdot 2! < 27 \quad \Rightarrow \quad 8 \cdot 2 < 27 \quad \Rightarrow \quad 16 < 27 \quad \text{(True)} \]
Therefore, the inequality holds for \( n = 3 \).

4. **Test \( n = 4 \)**:
\[ 2^4 (4-1)! < 4^4 \]
This simplifies to:
\[ 16 \cdot 3! < 256 \quad \Rightarrow \quad 16 \cdot 6 < 256 \quad \Rightarrow \quad 96 < 256 \quad \text{(True)} \]
Therefore, the inequality holds for \( n = 4 \).

5. **Test \( n = 5 \)**:
\[ 2^5 (5-1)! < 5^5 \]
This simplifies to:
\[ 32 \cdot 4! < 3125 \quad \Rightarrow \quad 32 \cdot 24 < 3125 \quad \Rightarrow \quad 768 < 3125 \quad \text{(True)} \]
Therefore, the inequality holds for \( n = 5 \).

6. **General Observation**:
From the tests, we observe that the inequality holds for \( n = 3, 4, 5 \) and likely for all \( n > 2 \).

### Conclusion:
The inequality \( 2^n (n-1)! < n^n \) holds true for all natural numbers \( n \) greater than 2.

Thus, the final answer is:
\[ n > 2 \]