Firstly, we note that

$$123^{321} \equiv 23^{321} = (23^4)^{80}\times 23^1\pmod{100}$$

The base of the first term, $23^4$, ends with $1$. Because its exponent, $80$, ends with $0$, we conclude the tens digit of $(23^4)^{80}$ must be $0$. In another word, $(23^4)^{80}\equiv 01 \pmod{100}$. Setting this to the above relation leads to $$123^{321}\equiv 1\times 23=\boxed{23}\pmod{100}$$