Proof using induction n! I wanna know how to proof using induction; I saw this example in a discrete book; however, i could not solve it; the question is below:
Prove, using induction, that $3^{n} < n!$ for all $n ≥ 7$
 A: Initial case: The case clearly holds for $n = 7$. (Easily checked with a calculator)
Inductive hypothesis: $3^{n-1} < (n-1)!$
Inductive step:
$$ 3^n = 3 \cdot 3^{n-1} < 3 (n-1)! < n (n-1)! = n!
$$
I think the purpose of this problem is to demonstrate how to use induction, instead of the importance of the fact $3^n < n!$.
A: I'm not sure what "induction" is ( i'm not an English native speaker ) but that's how I would do it :
First take the case of $n = 7$
$3^7= 2187$ and $7! = 5040 $
so it's true for $n=7$
Then suppose for some $n \geq 7$ that we have $3^n < n! $
now we multiply both sides of by 3 and get 
$3^{n+1} < 3(n!)$   
and we know $n \leq 7$ so 
$3(n!) < (n+1)! $   
So if we combine our two above steps we get:
$3^{n+1}< (n+1)!$
There you go! It's true for $n=7$ and  if true for any $n \geq 7$ then it's true for $n+1$ so you can deduce it's true for any $n \geq 7$.
A: First, show that this is true for $n=7$:
$3^7<7!$
Second, assume that this is true for $n$:
$3^n<n!$
Third, prove that this is true for $n+1$:
$3^{n+1}=$
$3\cdot\color\red{3^n}<$
$3\cdot\color\red{n!}<$
$7\cdot{n!}<$
$(n+1)\cdot{n!}=$
$(n+1)!$

Please note that the assumption is used only in the part marked red.
