Sum of Factorials 
What is the hundreds digit when 2014! + 2013! + ... + 3! + 2! + 1! is expressed as an integer?

I was hoping to find out some sort of pattern by trying the first few factorials, but far as I can see there's none. Also, there seems to be some formula for this, but I'm not really sure if I understand them. How would I solve this? 
 A: Perform the computations modulo $1000$ (no calculator required):
$$\begin{align}1!&\equiv1,\\
2!&\equiv2,\\
3!&\equiv6,\\
4!&\equiv24,\\
5!&\equiv120,\\
6!&\equiv720,\\
7!&\equiv40,\\
8!&\equiv320,\\
9!&\equiv880,\\
10!&\equiv800,\\
11!&\equiv800,\\
12!&\equiv600,\\
13!&\equiv400,\\
14!&\equiv600,\\
15!&\equiv0.
\end{align}$$
Then the sum (easier backward)
$$313.$$
A: Notice that for $n\ge15$ the factorial will end upto $000$ ,
($15!$ or more contain $125=5^3$ as a factor and when this multiplied by $2^3$ will lead to $000$ at the end)
so we need to find hundreds digits of $14!+13!+12!+......+1!$ and this is equal to $313$ because the sum is $93928268313$ and so the hundreds digits of $$2014! + 2013! + ... + 3! + 2! + 1!$$is 313.
A: Hint:
Compute $15!$ and note the three last digits.
A: You can find the hundreds digit of a number if you know the remainder when divided by $1000$. Can you see that if $k$ divides $n_0!$ then $k$ divides $n! \, \, \forall n \geq n_0$?
If that's clear then note that $15!$ is divisible by $1000$ and you have your answer.
