<s>[INST] Which natural numbers satisfy $2^n > n^2$? 
Which natural numbers satisfy $2^n > n^2$ ?

My work. Step 1: $n = 1 $, $2^1 > 1^2$. True.
For $n = k$, $2^k > k^2$. For $n = k+1$,
$$ 2^{(k+1)} > (k+1)^2 \\
2\cdot 2^k>k^2+2k+1 \\ 
2^k+2^k > k^2+2k+1$$


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*$2^k > k^2 \text{ - from step 1}$

*$2^k > k^2+2k+1$
How I can find the numbers now? 
 [/INST] For $n = 1$ is true, but for $n = 2,3,4$ is not. but for $n \geq 5$ is true.  
Hence, the solution set is $\{1\}\cup[5,\infty)$.
Now we assume is true for $n = k$, that is
$2^k > k^2$, we will have to show that it is true for $n = k+1$.
$2^{k+1} = 2(2^k)$, but $2^k > k^2$ by induction hypothesis then $2^{k+1} > 2k^2 > k^2$.
Hence by the induction principle it is true for all $n \geq 5$ (also for $n = 1$).
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