4. The product of all natural numbers from 1 to $\mathrm{n}$ is denoted by $n$! (read as “ $n$ - factorial”). Which of the numbers is greater: 200! or $100^{200}$?
Solution. We will prove that $200!<100^{200}$, for which we will divide the factors on the left into quartets of the form $k(101-k)(100+k)(201-k)$ for $k=1,2, \ldots, 50$ and compare each quartet with $100^{4}$: $\left(k^{2}-101 k\right)\left(k^{2}-101 k-20100\right)<$ $100^{4}$. The variable $t=\left(k^{2}-101 k\right)$ for $k=1,2, \ldots, 50$ monotonically decreases from -100 to -2550, the inequality $t^{2}-20100 t-100^{4}<0$ is satisfied for such values of $t$: the quadratic trinomial here decreases and takes negative values at the endpoints.