Proving $(n+1)!>2^{n+3}$ for all $n\geq 5$ by induction I am stuck writing the body a PMI I have been working on for quite some time.
Theorem: $∀n∈N ≥ X$, $(n+1)!>2^{n+3}$
I will first verify that the hypothesis is true for at least one value of $n∈N$.
Consider $n=3$: (not valid)
$$(3+1)!>2^{3+3} \implies 4!>2^{6} \implies 24>64$$
Consider $n=4$: (not valid)
$$(4+1)!>2^{4+3} \implies 5!>2^{7} \implies 120>128$$
Consider $n=5$: (valid)
$$(5+1)!>2^{5+3} \implies 6!>2^{8} \implies 720>256$$
Consider $n=6$: (valid)
$$(6+1)!>2^{6+3} \implies 7!>2^{9} \implies 5040>512$$
So clearly $X$ is $5$.
For the inductive assumption, we will assume the hypothesis holes from $n=5$ up to some arbitrary values $k$: $(k+1)!>2^{k+3}$
This is where I am lost. Originally I had written:
Now I will prove true for $k+1$ showing that: $(k+2)!>2^{k+4}$. Consider the $k+1$ term:
$$(k+2)(k+1)k! = (k+2)(k+1)2^k = (k+2)(k+1)2^k = (2)(2)(2^k) = (4)(2^k) = 2^{k+4}$$
by the inductive assumption since $k>5$ so $k+1>2$ and $k+2>2$ but I know this isn't correct.
I know the proof should look something like this, but I have no idea why:
$$(k+2)(k+1)!>(k+2)(2^{k+3})>2(2^{k+3})=2^{k+4}$$

For $n\geq 5$, let $S(n)$ denote the statement
$$
S(n) : (n+1)! > 2^{n+3}.
$$
Base step $(n=5)$: $S(5)$ says $6!>2^8$, and this is true since $720>256$. 
Inductive step: Fix some $k\geq 5$ and suppose that 
$$
S(k) : (k+1)!>2^{k+3}
$$
is true. We must now prove that $S(k+1)$ follows where
$$
S(k+1) : (k+2)! > 2^{k+4}.
$$
Starting with the left-hand side of $S(k+1)$,
\begin{align}
(k+2)! &= (k+1)!\cdot(k+2)\tag{by definition}\\[0.5em]
&> 2^{k+3}(k+2)\tag{by $S(k)$}\\[0.5em]
&\geq 2^{k+3}\cdot 7\tag{since $k\geq 5$}\\[0.5em]
&> 2^{k+3}\cdot 2\\[0.5em]
&= 2^{k+4},
\end{align}
we see that the right-hand side of $S(k+1)$ follows. Thus, $S(k)\to S(k+1)$, completing the inductive step.
Thus, by mathematical induction, for all $n\geq 5$, the statement $S(n)$ is true. $\blacksquare$
