A Proof by induction 
So my base case is $n=4$
$$LHS= 2^n = 2^4 = 16$$
$$RHS= n! = 4! = 24$$
$$16<24$$
$$n=k$$
$$2^k < k!$$
$$n=k+1$$ 
$$2^{k+1} < (k+1)!$$
After this step I'm not sure how to solve the problem.
 A: First, show that this is true for $n=4$:
$2^4<4!$
Second, assume that this is true for $n$:
$2^n<n!$
Third, prove that this is true for $n+1$:
$2^{n+1}=\color\red{2^n}\cdot2<\color\red{n!}\cdot2<n!\cdot{n}<n!\cdot(n+1)=(n+1)!$

Please note that the assumption is used only in the part marked red.
A: First of all, this is not a good way to set out an inductive argument. Your proof of the base case is fine. But you should probably write something along the lines of "so the result holds for n=4" or something similar, just to make it clear that you have indeed shown that this case holds. 
Next, state your inductive hypothesis. In this case, that is "Assume that $2^k<k!$ for some $k\in\mathbb N$ with $k\geq 4$". As you have written it above, it is not clear at all what you are trying to show, or what you are assuming!
You finish the proof by an argument (see the other answers for a hint) to show that if the result holds for $k$ (as above), then it also holds for $k+1$. Finish the proof off by stating that the result holds for all $n\geq 4$ by induction.
A: Hint : $2^{k+1}=2\times 2^k$ , $(k+1)!=k!\times (k+1)$
You can assume $2^k<k!$ and you have to show $2^{k+1}<(k+1)!$
A: HINT:  $$2^{k+1}=2^k\cdot 2 < k!\cdot 2 < k!\cdot (k+1) = (k+1)! \Rightarrow 2^{k+1}<(k+1)!$$
A: As Peter mentioned, $2^{k+1} = 2 \times 2^k$ and $(k+1)! = k! \times (k+1)$. You know by your inductive hypothesis that $2^k < k!$ which means that....
