We are tasked with proving or investigating the inequality:

$n! \geq 2^{n-1}$

Step 1: Check small values of $n$

Let’s test the inequality for small values of $n$.

• For $n = 1$: $1! = 1 \quad \text{and} \quad 2^{1-1} = 2^0 = 1$ So, $1! = 2^0$, which satisfies the inequality $1 \geq 1$.

• For $n = 2$: $2! = 2 \quad \text{and} \quad 2^{2-1} = 2^1 = 2$ So, $2! = 2^1$, which satisfies $2 \geq 2$.

• For $n = 3$: $3! = 6 \quad \text{and} \quad 2^{3-1} = 2^2 = 4$ So, $6 \geq 4$, which satisfies the inequality.

• For $n = 4$: $4! = 24 \quad \text{and} \quad 2^{4-1} = 2^3 = 8$ So, $24 \geq 8$, which satisfies the inequality.

Thus, the inequality holds for $n = 1, 2, 3, 4$.

Step 2: Inductive Proof

Let's prove this inequality using mathematical induction.

Base case: $n = 1$

We already checked that for $n = 1$, the inequality holds: $1! = 1 \quad \text{and} \quad 2^{1-1} = 1$ So, $1 \geq 1$ is true.

Inductive Hypothesis

Assume the inequality holds for some $n = k$, i.e., $k! \geq 2^{k-1}$ We need to prove that it holds for $n = k+1$, i.e., $(k+1)! \geq 2^k$

Inductive Step

We know that $(k+1)! = (k+1) \times k!$. Using the inductive hypothesis, we get: $(k+1)! = (k+1) \times k! \geq (k+1) \times 2^{k-1}$ Now we need to show that $(k+1) \times 2^{k-1} \geq 2^k$, which simplifies to: $(k+1) \times 2^{k-1} \geq 2^k$ Dividing both sides by $2^{k-1}$, we get: $k+1 \geq 2$ This is true for all $k \geq 1$.

Thus, by the principle of mathematical induction, the inequality $n! \geq 2^{n-1}$ holds for all $n \geq 1$.

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Related Questions:

1. What are other examples of inequalities involving factorials and powers?
2. How can you use induction to prove other types of inequalities?
3. Why does factorial grow faster than exponential functions after a certain point?
4. Can we compare $n!$ and $3^n$? Which grows faster?
5. How