Prove by induction: $3^{n-2}\le(n-1)! : \forall n\ge 6$
The base case and hypothesis are trivial, we want to show that: $3^{n-1}\le(n)! : \forall n\ge 6$, but I get stuck very early: $3^{n-1}\le \frac{3\cdot3^{n-1}}{3}\overset{I.H}{\le}3(n-1)!\le {?}$
Any hints please?
Suppose $3^{n-2} \leq (n-1)!$. Then $$3^{n-1} = 3 \cdot 3^{n-2} \leq 3 \cdot (n-1)! \leq n \cdot (n-1)! = n!$$ as long as $n \geq 3$.