Prove that $3^n > 3n$ for integer $n\geq2$

Question

How would we prove, by contradiction that $3^n > 3n$ for integer $n\geq2$. I'm having trouble on where I should start in tackling this question.

I know that we should first state the negative of the statement, ie, we assume $3^n \leq 3n$ for integer $n\geq2$, but I am not sure how to prove this assumption wrong.

Answer 1

Let $3^n\leq3n$ for some $n\geq2$.

Thus, $$3^n=(1+2)^n\geq1+2n+\frac{n(n-1)}{2}\cdot4>3n,$$ which is contradiction.

Answer 2

For $n\ge1$, Bernoulli's Inequality gives $$ \begin{align} 3^{n-1} &=(1+2)^{n-1}\\ &\ge1+2(n-1)\\ &=2n-1\\ &\ge n\tag1 \end{align} $$ Multiply $(1)$ by $3$ yields $$ 3^n\ge3n\tag2 $$