Factorial Summation

In the above sum, we know that for any number $n≥10$ , $n!$ will have trailing zeroes $≥2$ .

Therefore, for finding last $2$ digits of the above sum, we have to find,

$\displaystyle \sum^{9}_{n=1}n!=409113 \Rightarrow \boxed{13}$

In a more stripped down solution to Akshat, I find only the last two digits (note that $10! = 3,628,800$ ), which, if expand for any $n ≤ 9$ , would be

$1+2+6+24+(20)+(20)+(40)+(20)+(80) = 213 \equiv (13)$

Where the brackets refer to modulo 100 of the respective number (LHS is $n!\mod 100$ )