For which nonnegative integers n is. n^2\leq n!?

For which nonnegative integers n is
${n}^{2}\le n!?$
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Nathaniel Kramer

Let us determine the possible integers for which ${n}^{2}\le n!$ by constructing a graph of the two functions $f\left(x\right)={x}^{2}$ and $g\left(x\right)=x!$
The intersection of the two graphs is then $\left(3.5624,12.6906\right)$. For values n larger than 3.5624, the inequality ${n}^{2}\le n!$ will then hold.
Since n is an integer:
$n>3$
Moreover, we note that the equality holds for $n=0\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}n=1$ as well, since $0\ne 1\ge 0={0}^{2}$ and $1\ne 1={1}^{2}$

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