quick question about inductive proof

**zpwnchen** · October 16th 2009, 06:26 PM

Prove that: $3^n < n!$ n is an integer greater than 6.

Basic step:
$P(7): 3^7 < 7!$
2187 < 5040

Inductive Steps:
hypothesis: $P(k): 3^k < k!$
conclusion: $P(k+1): 3^{k+1} < (k+1)!$

Proof: $3^(k+1) = 3.3^k < 3.k! < (k+1)k! = (k+1)!$

why is this one: $(k+1)k!$ equal to $(k+1)!$ ?

Is there any better way to prove it?

**Plato** · October 16th 2009, 07:10 PM

If you have more than one character in an exponent, set off the whole exponent in braces.
[tex]3^{(k+1)}[/tex] gives $3^{(k+1)}$ instead of $3^(k+1)$ .

**Defunkt** · October 17th 2009, 02:30 AM

Originally Posted by zpwnchen

Prove that: $3^n < n!$ n is an integer greater than 6.

Basic step:
$P(7): 3^7 < 7!$
2187 < 5040

Inductive Steps:
hypothesis: $P(k): 3^k < k!$
conclusion: $P(k+1): 3^(k+1) < (k+1)!$

Proof: $3^(k+1) = 3.3^k < 3.k! < (k+1)k! = (k+1)!$

why is this one: $(k+1)k!$ equal to $(k+1)!$ ?

Is there any better way to prove it?

$k! = 1\cdot 2 \cdot 3 \cdot ... \cdot k$
$(k+1)! = 1\cdot 2 \cdot 3 \cdot ... \cdot k \cdot (k+1)$
$(k+1)k! = (k+1)(1\cdot 2 \cdot 3 \cdot ... \cdot k) = 1\cdot 2 \cdot 3 \cdot ... \cdot k \cdot (k+1) = (k+1)!$

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