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
$(k+1)^{(k+1)-3} \ge (k+1)!$
 A: 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
$(k+1)^{(k+1)-3} \ge (k+1)!$
To prove the inductive step to be true note that $(k+1)!=(k+1)\cdot k!$
$$(k+1)^{(k+1)-3} \geq (k+1)!$$
$$(k+1)^{k-2} \geq k!\cdot (k+1)$$
Since $k+1$ is always positive because $k \geq 9$ we can divide both sides by $(k+1)$
$$(k+1)^{k-3} \geq k!$$
 q.e.d
A: 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 
$$\frac{n^{n-3}}{n!}\ge 1.$$
A: With the base case of $n = 9$ established, you want to show that if
$$ n^{n-3} \geq n! ~: ~n \in \Bbb{Z_{\geq 9}}\tag1 $$
then
$$ [n+1]^{[n+1]-3} \geq [n+1]! ~.\tag2 $$
As the inequality shifts from (1) above to (2) above, the LHS is multiplied by
$$\left(\frac{n+1}{n}\right)^{n-3} \times (n+1) \tag3 $$
while the RHS is multiplied by $$(n+1). \tag4 $$
Since the first factor in (3) above is always greater than $1$, and the 2nd factor in (3) above equals the factor in (4) above, the inequality in (2) above is implied by the inequality in (1) above.
