Series used in proof of irrationality of $ e^y $ for every rational $ y $ Following from the book "An Introduction to the Theory of Numbers" - Hardy & Wright I am having trouble with this proof.  The book uses a familiar proof for the irrationality of e and continues into some generalizations that lose me.
In the following statement where is the series coming from or how is the statement derived?
$ f = f(x) = \frac{x^n(1 - x)^n}{n!} = \frac{1}{n!} \displaystyle\sum\limits_{m=n}^{2n} c_mx^m $
I understand that given $ 0 < x < 1 $ results in
$ 0 < f(x) < \frac{1}{n!} $
but I become confused on . . .

Again $f(0)=0$ and $f^{(m)}(0)=0$ if $m < n$ or $m > 2n.$  But, if $n \leq m \leq 2n $,
$ f^{(m)}(0)=\frac{m!}{n!}c_m $
an integer.  Hence $f(x)$ and all its derivatives take integral values at $x=0.$  Since $f(1-x)=f(x),$ the same is true at $x=1.$

All wording kept intact!
The proof that follows actually makes sense when I take for granted the above.  I can't however take it for granted as these are, for me, the more important details.  So . . . 

For your first question, this follows from the binomial theorem
$$x^n(1-x)^n=x^n\sum_{m=0}^{n}{n\choose m}(-1)^mx^m=\sum_{m=0}^{n}{n\choose m}(-1)^{m}x^{m+n}=\sum_{m=n}^{2n}{n\choose m-n}(-1)^{m-n}x^m$$
where the last equality is from reindexing the sum. Then let $c_m={n\choose m-n}(-1)^m$, which is notably an integer. I'm not quite clear what your next question is.
