Show that $n!>3^n$ for $n \ge 7$
My attempt:
Let the statement $P_n$ say that $n!>3^n$.
Base Case
Let $n=7$, then $P_7$ says that $7! > 3^7 \implies 5040>2187$
Inductive Step
Fix $k \ge 7$ Assume that $P_k$ is true, that is, $k!>3^k$
Show that $P_{k+1}$ is true,
This is where I'm concerned I may be wrong. Am I allowed to do what follows? Or must I always build my proof from the inductive step?
$(k+1)>3^{k+1} \implies k!(k+1)>3^k \cdot 3$
By our inductive step, we know that $k!>3^k$. Also, $(k+1)>3$ because $k \ge 7$
~My attempted proof ends here. The texts answer is shown below:
The answer for the inductive step is as follows;
$(k+1)!=(k+1)k!>(k+1)3^k$ this next part I don't really get. Why 7? It's not proving anything for any values greater than 7. $\ge(7+1)3^k=8\cdot3^k>3\cdot3^k=3^{k+1}$
It is all logical and legitimate. Why 7? Just because the statement is not true for $n < 7$. Say, for $n = 6$, $6! = 720 < 729 = 3^6$.
Don't feel weird about what you consider to be correct. If you are asked to prove the same statement for $n \geq 10^{100}$, which is certainly true, the steps will be exactly the same.