1. Find the last three digits of $9^{100}-1$.
1. Ans: 000
$$
9^{100}-1=(1-10)^{100}-1=1-\binom{100}{1} 10^{1}+\binom{100}{2} 10^{2}-\cdots+\binom{100}{100} 10^{100}-1=1000 k
$$
for some integer $k$. Thus, the last three digits are 000 .