How can we prove that $k! \ge k^{k/2}$?

Question

I've tried considering the product of $k$ square roots of $k$ to no avail. Does the bound $k! \le k^{k/2}$ come from another popular bound?

Perhaps Stirlings?

Answer 1

Note that $$(k!)^2=[1\cdot k][2\cdot (k-1)][3\cdot (k-2)]\cdots [k\cdot 1].$$ Now use the fact that for any integer $i$ from $1$ to $k$ we have $i(k+1-i) \ge k$. This is true because the function $x(k+1-x)$ climbs steadily in the interval $0\le x\le \frac{k+1}{2}$, and then decreases in the interval $\frac{k+1}{2}\le x\le k+1$.

Thus $(k!)^2 \ge k^k$, which yields the desired result.