$2006^{2005}!$

$(2000+6)^{2005}$ = ${2005} \choose {0}$ $\times 6^{2006}+2000 (X)$ , that is other term will not affect last 2 digits. So last 2 digits of this will be same as last 2 digits of $6^{2006}$ , which can be got by cyclicity, that is 76

$\phi(100)=40$

$2006^{2005} \equiv 2006^5 \equiv 6^5 = 7776 \equiv \boxed{76} \pmod{100}$