Factorial lower bound: $n! \ge {\left(\frac n2\right)}^{\frac n2}$

Suppose first that $n$ is even, say $n=2m$. Then

$$n!=\underbrace{(2m)(2m-1)\ldots(m+1)}_{m\text{ factors}}m!\ge(2m)(2m-1)\ldots(m+1)>m^m=\left(\frac{n}2\right)^{n/2}\;.$$

Now suppose that $n=2m+1$. Then

$$n!=\underbrace{(2m+1)(2m)\ldots(m+1)}_{m+1\text{ factors}}m!\ge(m+1)^{m+1}>\left(\frac{n}2\right)^{n/2}\;.$$

Hint: For a positive integers $a$, we have $a(n-a)\ge \frac{n}{2}$. This is because for $0\le x\le n$, the function $x(n-x)$ is increasing up to $x=\frac{n}{2}$, and then decreasing.

Pair the numbers $a$ and $n-a$ in the factorial.