Question: I am trying to show that the following sum is an integer:

$(\frac{a}{b} - (2 + \frac{1}{2!} + \frac{1}{3!} + ... + \frac{1}{n!}))n!$

where $a,b \in \mathbb{N}, b\neq 0$, and $n = \max(b,3)$.

I have no clue where to begin. Any hint would be appreciated.

Answer: 
This isn't much of a concrete proof but you can think about what happens when you distribute the $n!$. Firstly, $\frac{n!}{k!}$ will always be an integer if $k \leq n$ and they are both positive integers. Secondly, n is guaranteed to be greater than or equal to b, and therefore when you expand $n!$ into $1\*2\*3\*4\*...\*n$, you realize that b must be one of the integers and thus it must divide $n!$. Therefore since all of the components must be integers, the sum must be an integer.
As $k!\mid n!$ for $1 \leq k \leq n$, and $b \mid n!$, your sum is a integer.