Reading this: Simple proof that $\pi$ is irrational, I fail to understand the following part:
Since $n!f(x)$ has integral coefficients and terms in $x$ of degree not less than $n$, $f(x)$ and its derivatives (...) have integral values for $x=0$; also for $x=\pi=\frac{a}{b}$, since $f(x)=f(\frac{a}{b}-x)$
Assuming this, the rest I understand. But why is this true?
The derivatives $f^{(i)}(x)$ have constant term $0$ for $i<n$ since each term of $f(x)$ has degree at least $n$, and thus $f^{(i)}(0)=0$. For $i\ge n$, each term will have a multiplier of $i!$ in front of it, and $n!\mid i!$, so the constant term is an integer.
$n!f(x)$ is a polynomial with integer coefficients.
The derivative of a polynomial with integer coefficients is a polynomial with integer coefficents.
A polynomial with integer coefficients takes on integer values when evaluated at an integer.
