# quick question about inductive proof

• Oct 16th 2009, 06:26 PM
zpwnchen
Prove that: $\displaystyle 3^n < n!$ n is an integer greater than 6.

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

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

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

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

Is there any better way to prove it?
• Oct 16th 2009, 07:10 PM
Plato
If you have more than one character in an exponent, set off the whole exponent in braces.
$$3^{(k+1)}$$ gives $\displaystyle 3^{(k+1)}$ instead of $\displaystyle 3^(k+1)$.
• Oct 17th 2009, 02:30 AM
Defunkt
Quote:

Originally Posted by zpwnchen
Prove that: $\displaystyle 3^n < n!$ n is an integer greater than 6.

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

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

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

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

Is there any better way to prove it?

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