Tail of a huge power

Square repeatedly modulo $100$ and look for the exponent at which the power returns to $1$.

1. Modulo $100$ we have $17^2=289\equiv 89$, $17^4\equiv 89^2=7921\equiv 21$, $17^8\equiv 21^2=441\equiv 41$ and $17^{16}\equiv 41^2=1681\equiv 81$.
2. Then $17^{20}=17^{16}\times 17^4\equiv 81\times 21=1701\equiv 1$, so the powers of $17$ repeat with period dividing $20$.
3. Since $123456=20\times 6172+16$, we get $17^{123456}\equiv 17^{16}$ modulo $100$.
4. The last $2$ digits are $81$.

Answer: $81$