How to prove $4(n!)>2^{n+2}$ for $ n\geq 4$ with induction I've done the base step, but how do I prove it is true for $n+1$ without using a fallacy?
$$4(n!)>2^{n+2}\quad \text{for } n\geq 4$$
Please help.
 A: As AWertheim correctly noted, the claim you are asked to prove:

$$4(n!)>2^{n+2}\quad \text{for } n\geq 4$$

is equivalent to proving: 

$$ n! > 2^{n}\quad \text{for } n\geq 4\tag{1}$$  

$$\text{since}\quad 4(n!) > 2^{n+2} \iff 2^2\cdot n! > 2^2 \cdot 2^n   \iff n! > 2^n\quad \text{for } n\geq 4$$

$(1): P(n) \quad  n! > 2^n\quad \text{for } n\geq 4$
$P(4)$ I'll assume you've shown this to be true for the base case: $ 4! = 24 > 16 = 2^4 \quad \large \checkmark$
Assume the inductive hypothesis is true. $P(k):\quad$ There exist a $k \in \mathbb N, \;\; k\geq 4$ such that $\;k! > 2^k$. 
We need to show that $P(k+1)$ is true: $\quad(k+1)! > 2^{k+1}$. 
$$(k+1)! \;\; = \;\; \underbrace{(k+1)\,k! \;\;>\;\; \color{blue}{\bf (k+1)}2^{k}}_{P(k)}\;\; > \;\;\color{blue}{\bf 2}\cdot2^{k} \;\;= \;\;2^{k+1}$$ 
as desired. Note that we make use of the fact that $\color{blue}{\bf k+1 > 2}\;$ for all $\,k\geq 4$.
A: Hint: This is the same as proving $n! > 2^{n}$ by induction. To see how to do this, try writing out each side for some $n \geq 4$...
For instance for $n = 5$, we have:
$1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 > 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$
A term by term comparison should give you the intuition for a formal proof. 
A: Hint: If $n\geq 4$, then $(n+1)>2$, and for any positive numbers $a,b,c,d$, we have
$$a>b,\quad c>d\implies ac>bd.$$
