Induction: $n + 3 < n!$ for all $n>3$ I have a proof that I am trying to prove and I am getting stuck at the inductive hypothesis. This is my theorem:

For all real numbers $n>3$, the following is true: $n + 3 < n!$.

I have proven true for $n = 4$, and will assume true for some arbitrary value $k$, i.e.,
$$k + 3 < n!,$$
and I want to prove for $k+1$, i.e.,
$$(k+1) + 3 < (k+1)!.$$
Consider the $k+1$ term:
$$(k+1)+3 = ?$$
I am confused on how to approach the next step.
Ok here is how I am proceeding. It seems really long so if anyone has a better way let me know:
$$
=(k+3)+1
$$
$$
<(k!)+1
$$
$$
<k!+k!
$$
$$
=2k!
$$
$$
<(k+1)k!
$$
$$
=(k+1)!
$$
Therefore both sides are equivalent.
 A: Induction is overkill here.
For $n \gt 3$ we have that $n+3 \lt n + n = 2\cdot n \lt 1\cdot 2 \cdot3\cdots (n-1) \cdot n =n!$
A: As you are trying to solve this problem, I'll only give you a hint.
Inductive Step: we want to show $(n+1)+3 < (n+1)!$
That's equivalent to $n+4 < (n+1)\cdot n!$ by the property of the factorial.
We can distribute: $n+4 < (n\cdot n!) + (1\cdot n!)$
Can you take it from here?
A: Since by induction hypothesis,
$$k+3< k!$$
for $k>4$, multiply both sides by $(k+1)$ to get
$$(k+3)(k+1) < k! (k+1)$$
or
$$(k+3)(k+1) < (k+1)!$$
I'll leave the rest for you to think about, as a hint, remember that's an inequality.
A: Suppose $k! \gt k+3$ is true:

\begin{align*}
\ (k+1)! &=k!\cdot(k+1)
\\ &\gt(k+3)(k+1)
\\ &=k^2+4k+3
\\ &\gt k^2+k+3
\\ &\gt (k+1)+3\ldots(\text{where}\space k\gt 3)
\end{align*}

A: Suppose $n+3< n!$ and $(n+1)!\leq n+1+3$ , we see the following:
$$(n+1)!\leq n+1+3 \Rightarrow (n+1)n!\leq n+1+3 $$
$$\Rightarrow (n+1)(n+3)\leq(n+1)n!\leq n+1+3 $$
$$(n+1)(n+3)\leq (n+1)+3 \Rightarrow (n+1)(n+3)-(n+1) \leq 3$$
$$ \Rightarrow (n+1)(n+3-1)\leq 3$$
$$(n+1)(n+2)\leq 3$$
This should strike some thing s wrong... what is that?
