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I'm sharpening my math skills by myself since school never motivated me to do it, so right now I'm doing some induction exercises.

I just solved one and I'm not sure if it's right since it doesn't fit the solution I was given.

I skipped the base case.

$ 3^n > n^2 , n > 2 \\ 3^k > k^2 \\ 3^{k+1} > (k+1)^2 \\ 3^k > k^2 | \cdot 3 \\ 3^{k+1} > 3k^2 \\ 3k^2 > k^2 + 2k + 1 \\ 2k^2 > 2k+1 \\ 2k^2 - 2k > 1 \\ 2k(k+1) > 1 \\ \begin{cases}k > 2 => k+1 > 1 \\2k > 1\end{cases} \implies 3^{k+1} > (k+1)^2 $

The solution I saw was more bottoms-up and mine seems to be solved top-down anyway that's why I'm wondering.

Sorry for my Tex just learned some to hack this together. Thanks for taking time to read and answer my newbie doubt.

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    Your solution is fine. However, it lacks statements of what steps in induction you consider (known for $k$, needs to be proved for $k+1$ etc.), similarly: what is known and what needs to be shown. – 2017-02-04
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    The *proper* way to write an induction step for this would be $3^{k+1}\geq \dots >^{\text{induction hypothesis}}\dots\geq (k+1)^2$ in a single line. Starting with what you want to prove and arriving at a tautology is only mildly acceptable if you ensure that every step taken is reversible. In general proving that a statement implies a tautology is not enough to prove that the statement is true in general. – 2017-02-04
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    In this case, $3^{k+1}=3\cdot 3^k>^{\text{I.H.}}3\cdot k^2=k^2+k^2+k^2>^{\dagger}k^2+2k+1=(k+1)^2$ where I.H. is true by induction hypothesis and $\dagger$ is true by the fact that $k>2$. – 2017-02-04

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We can streamline the development as follows. First, we establish a base case. For $n=1$, this is $3^1>1^2$. Now, we assume that for some number $n\ge 2$ we have

$$3^n>n^2$$

Then, proceeding by induction and noting that for $n\ge 2$, $n^2\ge 2n>1$, we have

$$\begin{align} 3^{n+1}&=(3)\,3^n\\\\ &>3n^2\\\\ &=n^2+n^2+n^2\\\\ &>n^2+2n+1\\\\ &=(n+1)^2 \end{align}$$

And we are done!