jmc

Since for all $n\ge 10$, $n$ has two factors of 5, $n$ will end in two zeros, and so contributes nothing to the last two digits. So we need only compute $1!+2!+3!+4!+5!+6!+7!+8!+9!=1+2+6+24+120+720+5040+40320+362880$, and the relevant digits add up to $1+2+6+24+20+20+40+20+80$ whose last two digits are $\boxed{13}$.