The last four digits

By Binomial Theorem: $(100+1)^{999}\equiv \binom{999}{1}\cdot 100\cdot 1^{998}+1^{999}\equiv 99901\equiv 9901\pmod{10000}$ .

See pattern. 101 x 101 last 4 digits are 0201. 101 x 101 x 101 last 4 digits are 0301... thus 101^999 would be 9901 :)