23. $[\mathbf{1 2}]$ Find the smallest positive integer $n$ such that
$$
\underbrace{2^{2^{2^{\cdots}}}}_{n 2^{\prime} s}>\underbrace{((\cdots((100!)!)!\cdots)!)!}_{100 \text { factorials }} .
$$
Answer: 104
Note that $2^{2^{2^{2}}}>100^{2}$. We claim that $a>b^{2} \Longrightarrow 2^{a}>(b!)^{2}$, for $b>2$. This is because
$$
2^{a}>b^{2 b} \Longleftrightarrow a>2 b \log _{2}(b),
$$
and $\log _{2}(b)2$. Then since $b^{b}>b$ ! this bound works. Then
$$
\underbrace{\left(2^{2^{2 \cdots 2}}\right)}_{m 2^{\prime} \mathrm{s}}>\underbrace{((((100!)!)!)!\ldots)^{2}}_{m-4 \text { factorials }}
$$
for all $m \geq 4$ by induction. So $n=104$ works. The lower bound follows from the fact that $n!>2^{n}$ for $n>3$, and since $100>2^{2^{2}}$, we have
$$
\underbrace{(((100!)!)!)!\ldots)}_{100 \text { factorials }}>\underbrace{2^{2^{2^{100}}}}_{1002^{\prime} \mathrm{s}}>\underbrace{2^{2^{2}}}_{1032^{\prime} \mathrm{s}} .
$$