Induction for $3^n\geq n^2$ I need to prove by induction that $3^n\geq n^2$
What I have so far:


*

*Basis: $n = 1 \rightarrow 3^1 \geq 1^2 \rightarrow 3 \geq 1$

*Condition: $3^n\geq n^2$

*Assumption: $3^{n+1}\geq (n+1)^2$

*Basis: $3^n\geq n^2$
$3^{n+1}\geq (n+1)^2 
= 3(3^n) \geq n^2+2n+1 $
That's where I got stuck. I have no idea how to continue from there. 

We verify that it holds for $n=2$ since $3^2\ge2^2$
We assume that it holds for $n:$ $3^n\ge n^2(*)$ and we want to show that $3^{n+1}\ge(n+1)^2$
From $(*)$ we have by multiplying with $3$:
$3^{n+1}\ge3n^2$ 
But
$3n^2\ge(n+1)^2$ since $3n^2-(n+1)^2=2n^2-2n-2=n^2-2+(n-1)^2\ge0$
Since $n^2-2\ge0, (n-1)^2\gt0$ $\forall n\ge2$
