<s>[INST] Use induction to prove that the inequality, $n! < \left(\frac{n}{2}\right)^n$ continues to hold after $n=6$. I am confused how you would continue with the inductive step. Thanks
 [/INST] Basis step,
$P_6 : 6!<(\frac{6}{2})^6\rightarrow 720<3^6=729$. True 
Inductive step, suppose
$P_n$ is true, prove that $P_{n+1}$ is also true.
Note $P_n:n!<(\frac{n}{2})^n$ and $P_{n+1}:(n+1)!<(\frac{n+1}{2})^{n+1}$
Observe that, 
$$\begin{align}(\frac{n+1}{2})^{n+1}&=(\frac{n}{2}+\frac{1}{2})^{n+1} \\&>(\frac{n}{2})^{n+1}+(n+1)\cdot\frac{n^n}{2^{n+1}}+\frac{n(n+1)}{2}\cdot(\frac{n^{n-1}}{2^{n+1}})\tag{binomial}\\&=\frac{(5n+3)}{4}\cdot(\frac{n}{2})^n\\&=(n+\frac{n+3}{4})\cdot(\frac{n}{2})^n\\&>(n+1)\cdot(\frac{n}{2})^n,\forall n\ge6\end{align}$$
So, start with $P_n$,
$$\begin{align}n!(n+1)&<(\frac{n}{2})^n\cdot(n+1)\\&<(\frac{n+1}{2})^{n+1}\end{align}$$
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