Understanding $3^n < n!$

Question

In my class, we are given the answer to this proof. I understand how the inequality was simplified, but don't understand why the bolded statement is true for $k+1,$ or more simply, how that proves by MI that $3^n < n!$ when $n > 6.$

Prove that $3^n < n!$ if $n$ is an integer greater than $6.$

Sol: The basis step is $n = 7,$ and indeed $3^7 < 7!,$ since $2187 < 5040.$

Assume that the statement is true for $k,$ where $k > 6.$

Then $\mathbf{3^{k+1} = 3 \cdot 3^k < (k + 1) \cdot k! = (k + 1)!}.$ Hence, the statement is true for $k + 1,$ where $k > 6.$

Answer 1

This is proof by induction.

We denote $3^n < n!$ by $P(n)$, with $n > 6$. First, we check for an arbitary $n > 6$ if $P(n)$ holds. Let $n = 7$. Obviously, $3^7 < 7!$ (check). Now we assume $P(n)$ holds.

$P(n+1)$: $3^{n+1}= 3^n*3 < (n+1)! =n!*(n+1)$. Obviously, we know $3 < (n+1)$ for $6<n$. Thus, $P(n) \Rightarrow P(n+1)$. Q.e.d.

Answer 2

"Assume that the statement is true for $k,$ where $k > 6.$" This assumption says that $3^k < k!$ if $k$ is an integer greater than $6.$ It also says $k > 6.$ Therefore the assumption implies that $3^k < k!.$

Moreover, $k > 6$ implies that $3 < k + 1.$

So the assumption gives us $3 < k + 1$ and $3^k < k!,$ which together imply that $3 \cdot 3^k < (k + 1) \cdot k!.$