I don't know how to go on from here. This is what I have done so far minus the base case which is trivial.
Prove by induction the inequality:
$n^{n-3}\geq n!$ for $n\geq 9$
the hypothesis:
$j^{j-3}\geq j!$ for $j\geq 9$
the claim wit the induction step $(j+1)$:
$(j+1)^{(j+1)-3}\geq (j+1)!=j!(j+1)$ for $j\geq 9$
steps so far:
$(j+1)^{(j+1)-3}\Rightarrow$
$(j+1)^{(j-2)}\Rightarrow$
$\frac{(j+1)^{j}}{(j+1)^{2}}\Rightarrow$
