How can I deduce that this is always an integer? I am currently trying an elementary proof of the irrationality of $e$, which is defined as the unique positive integer such that $\int_{1}^{e} \frac{dx}{x}=1$.
The idea is to define $$S_n = \sum_{k=0}^{n} \frac{1}{k!}$$
After a little work, one may deduce that $$0 < \left (e-S_n \right )n! < \frac{3}{n+1}$$
and therefore $\left ( e-S_n \right )n!$ is never an integer for $n\geq 2$.
The idea now is to suppose $e=\frac{p}{q}$ for coprime integers $p$ and $q$ and to show (by way of contradiction) that $\left ( e-S_n \right )n!$ is an integer for all $n \geq 2$.
Now from the definition of $S_n$ and with a little algebra, we have
$$\left (e - S_n \right )n! = \frac{n!p}{q}- \sum_{k=0}^{n-1} \left ( n-k \right )!$$
Now the right hand sum is clearly an integer, so all that remains to do is to deduce that $\frac{n!p}{q}$ is an integer. But this is not at all obvious to me ie: $(n,q)=(3,5)$ does not yield an integer. Of course, if $n$ is sufficiently large then it will 'eventually' become an integer. But if the point is to prove that it is an integer for ALL $n \geq 2$, then it seems that we cannot arbitrarily increase $n$.
But mathematically speaking we should surely be able to obtain a contradiction here, so what is going on?

Your calculations about $(e-S_n)n!$ hold for every $n$. Rather than showing that $(e-S_n)n!$ is not an integer for all $n \geq 2$, you just need to show that $(e-S_n)n!$ is not an integer for some $n \geq 2$, contradicting what you proved above. Choose $n = q\cdot k$ for integer $k$, so that $\dfrac{n!p}{q}$ is an integer.
