Show that $n^{n-3} \ge n!$ for n=9, 10,... Show that $n^{n-3} \ge n!$ for n=9, 10,...

I have tried to n=9
$9^{9-3} = 9^6 = 531411$ 
$9! = 362880$ 
So $9^6 \ge 9!$ is true

My question is how do I prove it by every for n=9, 10,... by induction? I have tried the following, but need some hints and or corrections. 

Proof by induction
$ P(n): n^{n-3} \ge n!$ 
Base Case
$ n=9$ 
$9^6 \ge 9!$
Inductive Hypothesis
Assume P(k) is True
$k^{k-3} \ge k!$
Inductive Step
Show P(k+1) is True
$(k+1)^{(k+1)-3} \ge (k+1)!$

You don't really need induction. Indeed
$$\frac{n^{n-3}}{n!}=\frac1{n^3}\cdot\frac n1\cdot\frac n2\dotsm\frac n6\dotsm\frac nn=\frac{n^3}{6!}\cdot\frac n7\cdot\frac n8\dotsm1 $$
Now if $n\ge 9$, $\dfrac{n^3}{6!}\ge\dfrac{9^3}{6!}=\dfrac{729}{720} $. Hence all factors are $\ge 1$, so that 
$$\frac{n^{n-3}}{n!}\ge 1.$$
