In the article $\textit{A simple proof that } \pi \textit{ is irrational}$, see: https://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society/volume-53/issue-6/A-simple-proof-that-pi-is-irrational/bams/1183510788.full (the article consists of exactly one page, so the preview is enough), the following two real functions are defined:
$\begin{align} f(x) &= \frac{x^n(a-bx)^n}{n!} \\ F(x) &= f(x) -f^{(2)}(x) +f^{(4)}(x) -...+(-1)^{n}f^{(2n)}(x) \end{align}$
with $n$ an integer that can be chosen later on in the proof. It is assumed that $\pi = \frac{a}{b}$ is a rational number. It is claimed in the article that $F(\pi) + F(0)$ is an integer, because $f^{(j)}(\pi)$ and $f^{(j)}(0)$ are integers. This seems fairly intuitive, but I'm having difficulties proving this fact. I tried using induction together with the definition of the derivative, but I get stuck quite early. Is anybody able to show me why this is the case? Thank you in advance!
This is a general result for polynomials with rational coefficients
$f$ is a polynomial having rational coefficients. Hence all its derivatives are polynomials with rational coefficients.
As the value of a polynomial with rational coefficients at a rational number is a rational number, we get the desired result.