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:
$$(k+2)(k+1)!>(k+2)(2^{k+3})>2(2^{k+3})=2^{k+4}$$

Your first two steps are correct. You have established that:$$(n+1)!\gt2^{n+3}$$is first true when $n=5$. You then correctly assumed that it holds for some $n=k\ge5$, giving you:$$(k+1)!\gt2^{k+3}\tag{1}$$And, finally you also proceeded in the right direction by then saying that you now need to prove that this implies it is also true for $n=k+1$. If we look at $n=k+1$ we get:$$(k+2)!=(k+2)(k+1)!$$Now use the result from (1) to deduce that:$$(k+2)!\gt(k+2)2^{k+3}$$You can then show that:$$(k+2)2^{k+3}=k.2^{k+3}+2.2^{k+3}=k.2^{k+3}+2^{k+4}\gt2^{k+4}$$
