If $n$ and $k$ are positive integers, then

$$
(n+1) \cdot(n+2) \cdot \ldots \cdot(n+k)<(n+k)^{k}
$$

Use this to determine which of the two numbers below is greater than the other:

$(100!)$ ! and $99!^{100!} \cdot 100!^{99!}$.
Solution

If $a$ is an arbitrary integer, using the hint in the statement and multiplying the following 100 inequalities

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

we have:

$$
(100 a)!<\left(100!\cdot a^{100}\right)^{a}
$$

It now suffices to substitute $a=99!$ to obtain

$$
\begin{aligned}
(100!)! & <\left(100!\cdot(99!)^{100}\right)^{99!} \\
& =(100!)^{99!} \cdot(99!)^{100!}
\end{aligned}
$$