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 $\log _{2}(b)<b^{2} / 2$ for $b>2$. 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 $$\underbrace{(((100!)!)!)!\ldots)}_{100 \text { factorials }}>\underbrace{2^{2 \cdots^{2^{100}}}}_{1002^{\prime} \mathrm{s}}>\underbrace{2^{2} \cdots^{2}}_{103}$$