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 . . . 
 A: I see 8 assertions in the highlighted region - are all 8 worrying you? 


*

*$f(0)=0$. I hope that one is no problem. 

*$f^{(m)}(0)=0$ for $m\lt n$. We have $f(x)=x^ng(x)$ where $g$ is just a polynomial, so $f'(x)$ has $x^{n-1}$ as a factor, $f''(x)$ has $x^{n-2}$ as a factor, dotdotdot, $f^{(m)}(x)$ has $x^{n-m}$ as a factor, and is thus zero at $x=0$. OK? 

*$f^{(m)}(0)=0$ for $m\gt2n$. Well, $f$ is a polynomial of degree $2n$, so any derivative beyond $2n$ is identically zero. 

*If $n\le m\le2n$ then $f^{(m)}(0)=m!c_m/n!$. If you differentiate $\sum_{j=n}^{2n}c_jx^j$ $m$ times, the value at zero is the constant term, and the constant term is the one that comes from differentiating the $c_mx^m$ term, which becomes $m!c_m$. OK? 

*$m!c_m/n!$ is an integer. Well, $c_m$ is a coefficient in $(1-x)^n$, so it is certainly an integer, and $m\ge n$, so $m!/n!$ is certainly an integer. OK?

*$f$ and its derivatives take integer values at zero. Well, that's what we've just established, right? 

*"Since $f(x-1)=f(x)$, .... This should be $f(1-x)=f(x)$, which you can verify by just replacing $x$ in the formula for $f$ with $1-x$ and seeing what happens. 

*"The same is true at $x=1$." Apply $d/dx$ any number of times to $f(1-x)=f(x)$ and then put in $x=0$ and you'll see the relation between the values and derivatives at $x=0$ and at $x=1$. 
A: 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.
