Im supposed to prove that $n! \leq (\frac{n}{2})^n$ by induction and I got to know that its only valid for $ 6 \leq n$.
I tried it solving this way: $6! \leq 3^6 \implies 720 \leq 729 \qquad \text{true}$
\begin{align}
(n+1)! &\leq (\frac{n+1}{2})^{n+1} \\
(n+1)(n)! &\leq (\frac{n+1}{2})^{n+1} \\
(n+1)\left( \frac{n}2
\right)^n &\leq (\frac{n+1}{2})^{n+1}
\end{align}
Stopped here because I have no idea how to solve this(wanted to induct from $n+1$ back to $n$ somehow, so getting like $n! \leq (\frac{n}{2})^n$ , but didnt work out).
Can you pls give me a hint how to solve this?
The key part of this problem is to realize $(1+{1\over n})^n\geq 2$.
We use the binomial theorem here: $(1+{1\over n})^n\geq 1+n\cdot1\cdot{1\over n}=2$
Now we apply our induction step:
$(\frac{n+1}{2})^{n+1}=(\frac{n}{2})^{n}\cdot({1+{1\over n}})^n\cdot({n+1\over2})\geq n!\cdot2\cdot ({n+1\over2})=(n+1)!$