Example 4 Find the last three digits of $2013^{2013^{2013}}$.

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Notice,
$$
\begin{array}{l}
2013 \equiv 3(\bmod 5), 2013^{2} \equiv-1(\bmod 5), \\
2013^{4} \equiv 1(\bmod 5), 2013 \equiv 13(\bmod 25), \\
2013^{2} \equiv-6(\bmod 25), 2013^{4} \equiv 11(\bmod 25), \\
2013^{8} \equiv-4(\bmod 25), 2013^{12} \equiv 6(\bmod 25), \\
2013^{20} \equiv 1(\bmod 25) .
\end{array}
$$

Since $2013 \equiv 1(\bmod 4), 2013^{20} \equiv 1(\bmod 4)$, therefore, $2013^{20} \equiv 1(\bmod 100)$.
$$
\begin{array}{l}
\text { Hence } 2013^{2013} \equiv 2013^{20 \times 100+13} \equiv 2013^{13} \equiv 13^{13} \\
\equiv(10+3)^{13} \equiv 10 \mathrm{C}_{13}^{12} \cdot 3^{12}+3^{13} \\
\equiv 3^{12} \times 133 \equiv(80+1)^{3} \times 33 \equiv 241 \times 33 \\
\equiv 41 \times 33 \equiv 53(\bmod 100) .
\end{array}
$$
$$
\begin{array}{l}
\text { By } 2013^{20} \equiv 13^{20} \equiv(10+3)^{20} \\
\equiv \mathrm{C}_{20}^{19} \cdot 10 \times 3^{19}+3^{20} \equiv 3^{19} \times 203 \\
\equiv\left(3^{5}\right)^{3} \times 3^{4} \times 78 \\
\equiv-7^{3} \times 81 \times 78 \equiv 51(\bmod 125),
\end{array}
$$

we get $2013^{40} \equiv(1+50)^{2} \equiv 101(\bmod 125)$,
$$
2013^{80} \equiv(1+100)^{2} \equiv-49(\bmod 125) \text {. }
$$

Hence $2013^{100} \equiv 1(\bmod 125)$.
$$
\text { By } 2013 \equiv 5(\bmod 8), 2013^{2} \equiv 1(\bmod 8) \text {, }
$$

we get $2013^{100} \equiv(\bmod 1000)$.
From $2013^{2013}=100 k+53$, we have
$$
\begin{array}{l}
2013^{2013^{2013}} \equiv 2013^{100 k+53} \equiv 2013^{53} \\
\equiv(2000+13)^{53} \equiv 13^{53} \equiv(10+3)^{53} \\
\equiv C_{53}^{2} \cdot 10^{2} \times 3^{51}+C_{58}^{1} \cdot 10 \times 3^{52}+3^{53} \\
\equiv 3^{51} \times 399(\bmod 1000) \\
\text { Also } 3^{7} \equiv 187(\bmod 1000), \\
3^{10} \equiv 187 \times 27 \equiv 49(\bmod 1000), \\
3^{17} \equiv 187 \times 49 \equiv 163(\bmod 1000), \\
3^{51} \equiv 163^{3} \equiv 747(\bmod 1000),
\end{array}
$$

Therefore, $3^{51} \times 399 \equiv 53(\bmod 1000)$.
In summary, the last three digits are 053.