<s>[INST] Proving by induction that $n! < (\frac{n+1}{2})^n$ As an analysis homework I have to prove by induction that
$n! < (\frac{n+1}{2})^n : (2 \le n \in\mathbb{N})$
For $n = 2$ this is trivial, but for $n+1$ no matter how I transform the equation I can't seem to get $(\frac{n+2}{2})^{n+1}$ on the right-hand side. I'm sure that this is an easy homework and I'm missing something blatantly obvious but the closest I got was $(n+1)! < \frac{(n+2)^{n+1}}{2^{n}}$.
Any help is greatly appreciated.
 [/INST] Without induction
and completely elementary:
$n!
= \prod_{k=1}^n k
$
so
$\begin{array}\\
n!^2
&= (\prod_{k=1}^n k)^2\\
&= (\prod_{k=1}^n k)(\prod_{k=1}^n k)\\
&= (\prod_{k=1}^n k)(\prod_{k=1}^n (n+1-k))\\
&= \prod_{k=1}^n k(n+1-k)\\
&= \prod_{k=1}^n (k(n+1)-k^2)\\
&= \prod_{k=1}^n \left(\dfrac{(n+1)^2}{4}-\dfrac{(n+1)^2}{4}+k(n+1)-k^2\right)\\
&= \prod_{k=1}^n \left(\dfrac{(n+1)^2}{4}-\left(\dfrac{(n+1)^2}{4}-k(n+1)+k^2\right)\right)\\
&= \prod_{k=1}^n \left(\dfrac{(n+1)^2}{4}-\left(\dfrac{n+1}{2}-k\right)^2\right)\\
&> \prod_{k=1}^n \left(\dfrac{(n+1)^2}{4}\right)\\
&= \left(\dfrac{(n+1)^2}{4}\right)^n\\
&= \left(\dfrac{n+1}{2}\right)^{2n}\\
\text{so}\\
n!
&> \left(\dfrac{n+1}{2}\right)^{n}\\
\end{array}
$
</s>