<s>[INST] $n! \leq \left( \frac{n+1}{2} \right)^n$ via induction I have to show $n! \leq \left( \frac{n+1}{2} \right)^n$ via induction.
This is where I am stuck:
$$\left( \frac{n+2}{2} \right)^{n+1}
 \geq \dots \geq
=2  \left( \frac{n+1}{2} \right)^{n+1}
= \left( \frac{n+1}{2} \right)^n(n+1)
\geq n!(n+1)
= (n+1)! $$
I approached this from both sides and this is the closest I can get. I realize that $n+2$ on the left has to be bigger than $n+1$ on the right, but I do not know who to show that it overpowers the factor two I have from the right.
What could I do to fill the dots? Currently, I just have it without the dots, but I would be happier if I could back it up.
 [/INST] Assuming $n! \le \left( \frac{n+1}{2} \right)^n$ is true, carry the induction step
$$
  (n+1) n!\leq (n+1) \left(\frac{n+1}{2}\right)^n =2
   \left(\frac{n+1}{2}\right)^{n+1} \stackrel{?}{\leq} \left(\frac{n+2}{2}\right)^{n+1}
$$
But the last inequality is just
$$
   2 \le \left( \frac{n+2}{n+1} \right)^{n+1} = \left( 1 + \frac{1}{n+1} \right)^{n+1} 
$$
It follows because:
$$
\left( 1 + \frac{1}{n+1} \right)^{n+1} = 
   \sum_{k=0}^{n+1} \binom{n+1}{k} \frac{1}{(n+1)^k} \ge
   \sum_{k=0}^{1} \binom{n+1}{k} \frac{1}{(n+1)^k} = 1 + (n+1) \frac{1}{n+1} = 2

$$
</s>