Answer
By mathematical induction, the formula is true for any integer $n\geq 6.$
Work Step by Step
$1!=1\leq 1^{3}$
$2!=2\leq 2^{3}$
$3!=6\leq 3^{3}$
$4!=24\leq 4^{3}$
$5!=120\leq 5^{3}$
$ 6!=720\gt 216=6^{3}$
So, we now try to prove the statement for all $n\geq 6$
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$1. \ (\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e})$ Check that the formula holds for $n=6$.
We just did this. Yes, it is true when $n=6.$
$2. \ (\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{p})$ Prove that IF the formula holds for any positive integer $n=k\geq 6$, THEN it also holds for the next integer, $n=k+1$.
Let us assume that for $n=k\geq 6$,
$ k!\gt k^{3}$
(we call this the step hypothesis)
For $n=k+1$,
the LHS equals $(k+1)!,\quad $
which we rewrite as
$LHS=k!\cdot(k+1)\qquad $
apply the hypothesis on $k!$
$LHS\gt k^{3}(k+1)$
Now, if we can prove that $k^{3}\gt(k+1)^{2}$ for $k\geq 6$,
we would have $LHS\gt(k+1)^{3}$, which would prove the statement.
So, we look at $(k+1)^{2}$
$(k+1)^{2}=k^{2}+2k+1$
$k$ is greater than both 1 and 2, so we can write
$(k+1)^{2}\lt k^{2}+k\cdot k+k^{2}=3k^{2}$
$k$ is greater than $ 3, $ so
$(k+1)^{2}\lt k\cdot k^{2}=k^{3}$
In other words, $k^{3}\gt(k+1)^{2}$
which means that the statement is true for $n=k+1.$
By mathematical induction, the formula is true for any integer $n\geq 6.$
