Answer
By mathematical induction, the formula is true for any integer $n\geq 5.$
Work Step by Step
$2^{1}\gt 1^{2}$
$2^{2}=2^{2}$
$2^{3}=8\leq 9=3^{2}$
$2^{4}=16=4^{2}$
$2^{5}=32\gt 25=5^{2}$
$2^{6}=64\gt 6^{2}$
$2^{7}=128\gt 7^{2} \quad ...$
So, we want to prove the statement for $n\geq 5$
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$1. \ (\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e})$ Check that the formula holds for $n=5$.
We just did this. Yes, it is true when $n=5.$
$2. \ (\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{p})$ Prove that IF the formula holds for any positive integer $n=k\geq 5$, THEN it also holds for the next integer, $n=k+1$.
Let us assume that for $n=k\geq 5$,
$ 2^{k}\gt k^{2}$
(we call this the step hypothesis)
For $n=k+1$,
the LHS equals$\quad 2^{k+1}$, which we rewrite as
$LHS=(2^{k})\cdot 2\quad $
and we apply the hypothesis
$2^{k+1} \gt k^{2}\cdot 2$
now, if $2k^{2}$ is greater than $(k+1)^{2}$, we are done (this would make the formula true for n=k+1).
$(k+1)^{2}=k^{2}+2k+1$
now, since $k\geq 5$, it follows that $2\displaystyle \lt\frac{k}{2}$ and $1\displaystyle \lt\frac{k^{2}}{2}$, so:
$(k+1)^{2} \lt k^{2}+\displaystyle \frac{k^{2}}{2}+\frac{k^{2}}{2}$
$(k+1)^{2} \lt 2k^{2}$
which makes the formula true for $n=k+1.$
By mathematical induction, the formula is true for any integer $n\geq 5.$
