Digits, Digits Everywhere

By AM-GM we have:

$\sqrt{(1000-x)x} \leq \dfrac{x+(1000-x)}{2} \implies x(1000-x) \leq 500^2$

$999!=999 \times 998 \times 997 \times \cdots \times 3 \times 2 \times 1=(999 \times 1) \times (998 \times 2) \times \cdots \times 500 \\ \cdots \leq (500^2)^{498} \times 500=500^{999}$

Note that equality hold in the first inequality iff $x=1000-x \implies x=500$ . This is not true in all the cases above so the inequality is strict:

$\color{#20A900}{\boxed{\boxed{999! <500^{999}}}}$

Here's an alternative solution, not using AM-GM, but using geometry! If we want to show that 500^999>999!, we can pair up opposite side terms in the factorial and show that: 500 * 500>999 * 1 and 500 * 500>998 * 2 and 500 * 500>997 * 3 and so on. More generally, we want to show that 500^2>(1000-k)*k

This could be done using AM-GM, or algebra, but instead you can consider this problem: Make a rectangle with perimeter 2000 that maximizes the area. Rectangles of this form have side lengths k and 1000-k, but we know that the maximum area is achieved by a square, with side lengths 500,500. Thus for all k not equal to 500, Area=(1000-k)k<2500 and we are done.

Start with $999!\;\boxed{?}\; 500^{999}$ where $\boxed{?}$ represents the unknown inequality operator.

Take the natural log of each side: $\ln(999!)\;\boxed{?}\;\ln(500^{999})=999\ln(500)$

The Stirling approximation states that for large $N$ we can approximate: $\ln(N!)\approx N\ln(\frac{N}{e})$ where $e\approx2.7$ is the mathematical constant

So we now have: $999\ln(\frac{999}{e})\;\boxed{?}\;999\ln(500)\Rightarrow\ln(\frac{999}{e})\;\boxed{?}\;\ln(500)$

Eliminate the logs: $\frac{999}{e}\;\boxed{?}\; 500$

Now since $e>2$ we trivially obtain: $\frac{999}{e}<500$

Meaning: $\boxed{999!< 500^{999}}$

$\prod_{i=1}^{999}i=\prod_{i=1}^{499}i(1000-i)\cdot500\leq\prod_{i=1}^{499}500^2\cdot500\mbox{ (AM-GM)}=500^{999}$

it is in the form $a!,500^a$ ;so We can check it for smaller numbers like $4!,500^4$ or $5!,500^5.$ So, $500^a$ is always bigger than $a!$ only if a is natural number.