Height 97 Find the last three digits of $2^{2^{2} 1100}$.

Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
$1000=2^{3} \times 5^{3}$ divided by the remainder.
Obviously, $x==1\left(\bmod ^{1} 2^{1}\right)$.
Next, consider the remainder of $x$ divided by $5^{3}$.
By Euler's theorem, $2^{2 / 5^{3}} \equiv \mathrm{I}\left(\bmod 5^{3}\right)$.

That is, $2^{+5^{2}}=1\left(\operatorname{mxl} 55^{3}\right)$.
Thus, $16^{35}=1\left(\bmod 5^{3}\right)$.
From equation (1), to find the remainder of $2^{1 \text { WNF }}$ divided by 25, by the binomial theorem we get
$$
\begin{array}{l}
2^{1} \text { (2)44 }=4^{454}=(5-1)^{4444} \\
\equiv(-1)^{x, 94}+\mathrm{C}_{494}^{1} \times 5 \times(-1)^{495} \\
=-115 \times 999=-115 \times(1000-1) \\
=-6+25 \times 2000=19(\bmod 25) \text {. } \\
\end{array}
$$

Therefore, $16 x=16^{256+24}=16^{211} \times\left(16^{25}\right)^{k}$
$$
\begin{array}{l}
=16^{24} \equiv(25-9)^{211} \equiv(\cdots 9)^{29} \equiv(10-1)^{20} \\
\equiv(-1)^{20}+C_{21}^{1} \times(-1)^{14} \times 10=1-200 \\
\equiv-199 \equiv 250-199 \equiv 51\left(\bmod 5^{3}\right) .
\end{array}
$$

That is, $16 . x=51(\bmod 125)$.
So, $16 x \equiv 51+125 \equiv 176=16 \times 11(\bmod 125)$. Hence $x \equiv 11(\bmod 125)$ (since 16 and 125 are coprime).
Where $k$ represents some positive integer.
From $x \equiv 0(\bmod 8)$ and $x \equiv 11(\bmod 125)$ we get
$$
\left\{\begin{array}{l}
125 x \equiv 0(\bmod 8 \times 125) \\
8 x \equiv 88(\bmod 8 \times 125)
\end{array} .\right.
$$

Therefore, $125 x-16 \times 8 x=0-16 \times 88(\bmod 1000)$,
That is, $-3 . x \equiv-1408(\bmod 1000)$.
$$
3 x \equiv 408 \cong 3 \times 136(\bmod 1000) \text {, }
$$

Hence $x \equiv 136(\bmod 1000)$ (since 3 and 1000 are coprime),
That is, the last three digits of $x=2^{2^{2} \text { (1001 }}$ are 136.