Prove By Mathematical Induction

I'm stuck on this question:

State the principle of mathematical induction, and use it to prove that for $\displaystyle n \geq 4$

$\displaystyle 4n! > 2^{n+2}$

So far I've done this:

$\displaystyle P(n)$ is a statement for each natural number $\displaystyle n$, where $\displaystyle n \geq 4$. $\displaystyle P(n)$ is true if:

- P(n) is true for n = 4.
- If it is true for arbitrary n, it is also true for n + 1.

For $\displaystyle n = 4$, $\displaystyle 4(4!) > 2^{6}$, $\displaystyle 96 > 64$ hence true.

For $\displaystyle n + 1$:

$\displaystyle 4(n + 1)! > 2^{n + 3}$

$\displaystyle (n + 1)! > 2^{n + 1}$

$\displaystyle n!(n+1) > 2^{n}(n+1)$ (Using hypothesis)

And that's where I get stuck. RHS should be $\displaystyle 2^{n+1}$, but that only occurs when n = 1.

Please help!