Prove that: $n! > (\frac{n}{2})^\frac{n}{2}$ for all $n \in \mathbb{N}$
Using induction:
Base case: $1! > (\frac{1}{2})^\frac{1}{2}$
Assume that $n! > (\frac{n}{2})^\frac{n}{2}$ for all $n \in \mathbb{N}$.
Now we have to show that $(n+1)! > (\frac{n+1}{2})^\frac{n+1}{2}$
$$(n+1)! = n! (n+1) > (n+1) \left(\frac{n}{2}\right)^\frac{n}{2}$$
I unfortunately stop there :(
It remains to show that $$(n+1) \left(\frac{n}{2}\right)^\frac{n}{2}\geq \left(\frac{n+1}{2}\right)^\frac{n+1}{2}$$ that is, after squaring, $$2(n+1)^2(n)^n\geq (n+1)^{n+1}$$ or $$2n+2\geq \left(1+\frac{1}{n}\right)^{n}.$$ Can you take it form here?