8. Find the last four digits of $7^{7^{-7}}$ (100 sevens).

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8. Solution: $\because 7 \equiv-1(\bmod 4), \therefore 7^{7^{7}} \equiv$ $-1(\bmod 4)(98$ sevens$)$, let $7^{\gamma^{\prime 7}}=4 k+3(98$ sevens$), k \in \mathbf{Z}^{+}$, then $7^{7^{\prime 7}}=7^{4 k+3}$ (99 sevens), $\because 7^{4} \equiv$ $1(\bmod 100), \therefore 7^{4 k} \equiv 1(\bmod 100), 7^{7^{\prime 2}} \equiv 7^{4 k+3}$ $\equiv 7^{3} \equiv 43(\bmod 100)$. Let $7^{\gamma^{7}}=7^{100 m+43}$ (100 sevens), $m \in \mathbf{Z}^{+}$.
$$
\because 7^{4} \equiv 2401(\bmod 10000), \therefore 7^{8} \equiv 4801
$$
$(\bmod 10000), 7^{16} \equiv 9601(\bmod 10000)$,
$$
7^{32} \equiv 9201(\bmod 10000), \quad 7^{64} \equiv 8401
$$
$(\bmod 10000), 7^{100} \equiv 7^{4} \times 7^{32} \times 7^{64} \equiv 2401 \times 9201$ $\times 8401 \equiv 1(\bmod 10000), 7^{100 m} \equiv 1(\bmod 10000)$, $7^{100 m+43} \equiv 7^{43} \equiv 7^{3} \times 7^{8} \times 7^{32} \equiv 2343$ $(\bmod 10000), \therefore 7^{\gamma^{7}}(100$ sevens$) \equiv 7^{100 m+43} \equiv$ $2343(\bmod 10000), 7^{7^{77}}(100$ sevens$)$ has the last four digits as 2343.