I was reading Bourbaki's proof that shows that $\pi$ is irrational, and the proof used the fact that if $\pi = \frac{a}{b}$ ($a$ and $b$ are positive integers) and $f(x)=\frac{x^n(a-bx)^n}{n!}$, then $\frac{d^kf}{dx^k}(0)$ and $\frac{d^kf}{dx^k}(\pi)$ are integers for $0\leq k \leq 2n$
My approach: $f(x) = f(\pi-x)$, so $(-1)^k\frac{d^kf}{dx^k}(\pi-x)=\frac{d^kf}{dx^k}(x )$, so proving that $\frac{d^kf}{dx^k}(0)$ is an integer is enough, but how do I prove this?
If $k<n$, $\frac{\mathrm d^kf}{\mathrm dx^k}(0)=0\in\mathbb Z$. And if $n\leqslant k\leqslant 2n$, since each monomial of the numerator of $f(x)$ is of the form $ax^m$, with $n\leqslant m\leqslant2n$, and since the denominator is $n!$, its $k$th derivative is an integer.