Prove n! > 2^n for n ≥ 4 by induction

Question

Prove by mathematical induction that $n! > 2^n$ for all integers $n \geq 4$ .

Solution — Step by Step

Claim: $P(n)$ : $n! > 2^n$ for all $n \geq 4$ .

Mathematical induction requires two steps:

Base case: Verify $P(n)$ holds for the starting value $n = 4$ .
Inductive step: Assume $P(k)$ is true for some $k \geq 4$ (inductive hypothesis), then prove $P(k+1)$ is also true.

LHS: $4! = 24$

RHS: $2^4 = 16$

Since $24 > 16$ , we have $4! > 2^4$ ✓

The base case holds.

Assume that $k! > 2^k$ for some $k \geq 4$ .

This is our inductive hypothesis — we take it as given for the purpose of this step and use it to prove the next case.

We need to show: $(k+1)! > 2^{k+1}$

Start with the left side:

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

Since $k \geq 4$ , we have $k+1 \geq 5 > 2$ . So:

(k+1) \times k! > 2 \times k!

Now apply the inductive hypothesis $k! > 2^k$ :

2 \times k! > 2 \times 2^k = 2^{k+1}

Combining these two inequalities:

(k+1)! = (k+1) \times k! > 2 \times k! > 2 \times 2^k = 2^{k+1}

Therefore: $(k+1)! > 2^{k+1}$ ✓

By the principle of mathematical induction:

The statement is true for $n = 4$ (base case).
If it is true for $n = k$ , it is also true for $n = k+1$ (inductive step).

Therefore, $n! > 2^n$ for all integers $n \geq 4$ . $\blacksquare$

Why This Works

The power of induction is the “domino effect.” Once you establish:

The first domino falls (base case)
Each domino knocking over the next (inductive step)

…you know ALL dominoes fall — without checking each one individually.

The key move in this proof: $(k+1)! = (k+1) \cdot k!$ . We use BOTH facts simultaneously:

$k+1 > 2$ (because $k \geq 4$ )
$k! > 2^k$ (the inductive hypothesis)

Multiplying two inequalities: if $a > b$ and $c > d$ (all positive), then $ac > bd$ .

So $(k+1) \cdot k! > 2 \cdot 2^k = 2^{k+1}$ .

Alternative Method — Verify the inequality intuitively

To understand WHY $n!$ grows faster than $2^n$ :

$n! = 1 \times 2 \times 3 \times \cdots \times n$ — factors increase as $n$ grows

$2^n = 2 \times 2 \times 2 \times \cdots \times 2$ — factors are always 2

For $n \geq 4$ : at position $k$ (for $k \geq 3$ ), the factor in $n!$ is $k \geq 3 > 2$ — larger than the corresponding factor in $2^n$ .

For $n = 4$ : $4! = 1 \times 2 \times 3 \times 4 = 24$ vs $2^4 = 2 \times 2 \times 2 \times 2 = 16$ . The first two factors are the same (1, 2). Then $3 > 2$ and $4 > 2$ in factorial — so factorial wins.

CBSE Class 11 induction proofs follow a strict format. Always write: (1) “Let $P(n)$ be…”, (2) “Verify $P(4)$ ”, (3) “Assume $P(k)$ is true: $k! > 2^k$ ”, (4) “To prove: $(k+1)! > 2^{k+1}$ ”, (5) “Proof:”, (6) “Hence by PMI, $P(n)$ is true for all $n \geq 4$ .” This structure earns full marks even if the middle step has minor errors.

Common Mistake

Students often forget to state WHY $(k+1) > 2$ in the inductive step. The fact that $k+1 \geq 5 > 2$ (since $k \geq 4$ ) is not trivial — it’s what allows us to write $(k+1) \cdot k! > 2 \cdot k!$ . Without this justification, the proof is incomplete. Always state: “Since $k \geq 4$ , $k+1 \geq 5 > 2$ .”

Question

Solution — Step by Step

Why This Works

Alternative Method — Verify the inequality intuitively

Common Mistake

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