## Thomas' Calculus 13th Edition

By mathematical induction, the formula is true for any integer $n\geq 6.$
$1!=1\leq 1^{3}$ $2!=2\leq 2^{3}$ $3!=6\leq 3^{3}$ $4!=24\leq 4^{3}$ $5!=120\leq 5^{3}$ $6!=720\gt 216=6^{3}$ So, we now try to prove the statement for all $n\geq 6$ --- $1. \ (\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e})$ 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 $n\geq 6.$