9.8. Which of the numbers is larger: (100!)! or $99!^{100!} \cdot 100!^{99!}$? (Recall that $n!=1 \cdot 2 \cdot \ldots \cdot n$.)
Answer. The second number is greater.

Solution. Let $a=99$ !. Then we need to compare the numbers $(100 a)!$ and $a^{100 a} \cdot(100 a)^{a}$. Notice that

$$
\begin{gathered}
1 \cdot 2 \cdot 3 \cdot \ldots \cdot a < a^{a} \\
(a+1)(a+2)(a+3) \cdot \ldots \cdot 2 a < (2 a)^{a} \\
(2 a+1)(2 a+2)(2 a+3) \cdot \ldots \cdot 3 a < (3 a)^{a} \\
\vdots \\
(99 a+1)(99 a+2)(99 a+3) \cdot \ldots \cdot 100 a < (100 a)^{a}
\end{gathered}
$$

Multiply these inequalities. On the left, we get the product of all numbers from 1 to $100 a$, i.e., exactly $(100a)! = (100!)!$, and on the right, the number

$a^{a}(2 a)^{a}(3 a)^{a} \cdot \ldots \cdot(100 a)^{a}=a^{100 a} \cdot(1 \cdot 2 \cdot 3 \cdot \ldots \cdot 100)^{a}=a^{100 a}(100!)^{a}$, i.e., exactly $a^{100 a}(100 a)^{a}=99!^{100!} \cdot 100!^{99!}$.

Comment. Only the answer - 0 points.

Proof of the inequality $(n!)!<(n-1)!^{n!} \cdot n!^{(n-1)!}$ for small values of $n$, not generalizable to the case $n=100-$ 0 points.

## 10th grade