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?