Prove using induction that $n^6 < 3^n$,for all $n > 18$

Question

Prove using induction (or using any other elementary precalculus techniques) that $$n^6 < 3^n, \forall n \geq 19.$$

I have no idea how to do this. Writing the induction step, I get that I need to prove that $$3n^6 > (n+1)^6,$$ and I don't know how to do so.

I want a proof that doesn't use calculus techniques.

Answer 1

Use that

$$3^{n+1}=3\cdot 3^n\stackrel{Ind. Hyp.}>3\cdot n^6 \stackrel{?}>(n+1)^6$$

and

$$3\cdot n^6 >(n+1)^6 \iff \frac{n+1}{n}<\sqrt[6] 3 \iff n>\frac1{\sqrt[6] 3-1}\approx 4.98$$

Answer 2

If $n\ge 19$,

$$(n+1)^6=n^6\left(1+\frac1n\right)^6\le3^n\cdot \left(1+\frac1{19}\right)^6<3^n\cdot1.1^6$$