Factorial lower bound: $n! \ge {\left(\frac n2\right)}^{\frac n2}$ A professor in class gave the following lower bound for the factorial
$$
n! \ge {\left(\frac n2\right)}^{\frac n2}
$$
but I don't know how he came up with this formula. The upper bound of $n^n$ was quite easy to understand. It makes sense. Can anyone explain why the formula above is the lower bound?
Any help is appreciated.

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