Prove that $n(n-1)<3^n$ for all $n≥2$ By induction Prove that $n(n-1)<3^n$ for all $n≥2$. By induction.
What I did: 
Step 1- Base case: 
Keep n=2
$2(2-1)<3^2$
$2<9$ Thus it holds.
Step 2- Hypothesis: 
Assume: $k(k-1)<3^k$
Step 3- Induction: 
We wish to prove that:
$(k+1)(k)$<$3^k.3^1$
We know that $k≥2$, so $k+1≥3$ 
Then $3k<3^k.3^1$
Therefore, $k<3^k$, which is true for all value of $n≥k≥2$
Is that right? Or the method is wrong? Is there any other methods?

I think that your solution is fine. However, I would phrase it slightly different.
Step-2. To be completely formal, I would say: Let $k>2$ and assume $k(k-1)<3^k$.
Step 3. We need to show $k(k+1)<3^{k+1}$. We have $$k(k+1)=k(k-1)+2k<3^k+2k<3^k+3^k+3^k=3^{k+1}$$
Where we have used the inductive hypothesis and the fact that $k<3^k$, which is true because $k>2$.
Notice that you can prove that the inequality is true for all $n\geq0$ (indeed the base case will become trivial!).
