Terminate quickly

For $n \ge 10$ we have that $n!$ has the factors $2,5$ and $10,$ and thus its last two digits will both be $0.$ Thus we just need to determine

$\left(\displaystyle\sum_{n=0}^{9} n!\right) \pmod{100} = \sum_{n=0}^{9} (n! \pmod{100})$

$\equiv (1 + 1 + 2 + 6 + 24 + 20 + 20 + 40 + 20 + 80) \pmod{100}$

$\equiv 214 \pmod{100} = \boxed{14}.$

The Factorial of all the numbers starting from 10 has the last two digits (atleast) as 0 and 0. This means we have to find the last two digits of

$\displaystyle \sum_{n=0}^9$

which is

$\displaystyle \sum_{n=0}^9 \pmod {100} \equiv 1 + 1 + 2 + 6 + 24 + 20 + 20 + 40 + 20 + 80 \pmod {100} \equiv 214 \pmod {100} = 14$