Proof by Induction that: $n!>3^n$ for $n>6$

Okay, I have this question: $n! > 3^n$, $6<n\in\mathbb Z$
Now, I solved it easily, but the answer is giving me trouble.
base - $n=7$, good.
assumption: $k! > 3^k$
Proof: $(k+1)! > 3^{k+1}$
Okay, played a little, got to this:
$k! \cdot (k+1) > 3^k \cdot 3$
lastly, got to:
$k>2$ Which gives me problem. I know k=n is greater than 6 (as said in the question).
But k is not greater then 2, it is greater than 6(not included).
Isnt this a contradiction?? I searched here for this solution as I got stuck, saw a few answers, others had $k>2$ the same as me and that is how they ended the question... Which is not a good answer, as I said, k is greater than 6, not than 2.... How am I continuing here from the solution I got?

EDIT: Okay, I saw people had trouble with my question. My question is: why the answer K>2 is good for this solution? since n>6 and k=n, so K should be also k>6. K=3,4,5,6 doesnt exist, but if I put k>2, it does exist.

Second Edit(Final): Managed to solve it, thanks to @Eric Towers.

Solution 1:

Have: $k! > 3^k$ for some particular $k > 6$.

Want: $(k+1)!>3k+1$

Proof:
$$ (k+1)! = (k+1)k! \\ > (k+1)3^k > (6+1)3^k\\ = (7)3^k > (3)3^k = 3^{k+1} \text{.} $$ Therefore, $(k+1)! > 3^{k+1}$, as desired.

At $(1)$, we use $k > 6$ so $k+1>6+1$, and then, since $3^k > 0$, $(k+1)3^k > (6+1)3^k$. At $(2)$, we use $7 > 3$ and $3^k > 0$, so $(7)3^k > (3)3^k$. In both cases, we use $a > b$ and $c > 0$ implies $ac > bc$.