Stuck on a step of a proof that $\pi$ is irrational. I'm trying to understand this proof that $\pi$ is irrational, yet I can't see why any derivative of the polynomial $$f(x)=\frac{x^n(a-bx)^n}{n!}$$
where $a,b\in\mathbb{Z}$ has integral values when $x=0$ or when $x=a/b$. I'd appreciate any help.

Edit: to be somewhat clearer, I fail to see why
$$f^{(i)}(a/b)\in \mathbb{Z}$$
for any natural number $i$.
 A: $f(x)=\frac{x^n(a-bx)^n}{n!}$. As Ivan Niven mentions in his paper
$$f(\frac{a}{b}-x)=\frac{\big(\tfrac{a}{b}-x\big)^n\big(a-b(\tfrac{a}{b}-x)\big)^n}{n!}=\frac{b^{-n}\big(a-bx\big)^n(bx)^n}{n!}=f(x)$$
Thus $f^{(j)}(x)=(-1)^jf^{(j)}(\tfrac{a}{b}-x)$ and so $f^{(j)}(0)=(-1)^nf^{(j)}(a/b)$.
Consequently, it suffices to check that each each $f^{(j)}(0)$ is an integer.
Using the binomial formula you obtain
$$ f(x)=\sum^n_{k=0}\frac{\binom{n}{k}}{n!}a^{n-k} b^{k}x^{n+k}=\sum^{2n}_{k=n}\frac{\binom{n}{k-n}}{n!}a^{2n-k}(-b)^{k-n}x^k$$
Recall the formula for power series for analytic functions around $0$:
$$\phi(x)=\sum^\infty_{k=0}\frac{\phi^{(k)}(0)}{k!}x^k$$
As a polynomial in $x$, $f(x)$ coincides with its Taylor series around $0$. Hence

*

*$f^{(k)}(0)=0$ for $1\leq k\leq n-1$ and $k>2n$

*$f^{(k)}(0)=\frac{k!}{n!}\binom{n}{k-n}a^{2n-k}(-b)^{k-n}$ for $n\leq k\leq 2n$. Each of which is an integer.

A: HINT.- An easy way to see clear your question is to consider your polynomial as a particular case of $f(x)=(P(x))^n$ (you have $P(x)=ax-bx^2$) Then you have
$$f'(x)=nP^{n-1}(x)P'(x)=P^{n-1}(x)Q_0(x)\\f''(x)=n(n-1)P^{n-2}(x)(P'(x))^2+nP^{n-1}(x)P''(x)=P^{n-1}(x)Q_1(x)\\f^{(3)}(x)=(n-1)P^{n-2}(x)P'(x)Q_1(x)+P^{n-1}(x)Q_1'(x)=P^{n-1}(x)Q_2(x)$$ and so on where the $Q_k(x)$ are succesively appearing polynomials. You can verify that all the derivatives have $P(x)$ as a factor excepting the $(nm)^{th}$ derivative which is equal to the constant $\pm (2n)!b$ (according to the parity of $n$), in your case the $(2n)^{th}$ derivative because of  the degree of $ax-bx^2$.
It follows that all your derivatives until the $(2n-1)^{th}$ when $x=0$ or $x=\dfrac ab$are equal to zero and the last one is equal to $\pm\dfrac{(2n)!b}{n!}$ which is in integer too.
A: The function $f(x)$ and all its derivatives are polynomials.
The value of a polynomial in $x$ at $x = 0$ is equal to the constant term.
We must have $n > 0,$ so the constant term of $f(x)$ is zero.
The constant term of the $k$th derivative of $f(x)$ is
$k! c_k,$ where $c_k$ is the coefficient of $x^k$ in $f(x).$
We know that $c_k = 0$ for $k < n$ (the term in $x^n$ is the non-zero term with the lowest power of $x$), so those derivatives are all zero, an integer.
For $k \geq n,$ the coefficient $c_k$ in $f(x)$ is some integer divided by $n!.$
But $k!$ is a multiple of $n!,$ so $k!c_k$ is an integer.
That's the proof for $x = 0,$ which you apparently already understand.

For $x = a/b,$ the observation is that $f\left(\frac ab - x\right) = f(x).$
Just plug $x \to \frac ab - x$ into the formula for $f(x)$ and watch as you get $f(x)$ back again.
Then
$$ 
f'(x) = \frac{\mathrm d}{\mathrm dx} f(x) =
\frac{\mathrm d}{\mathrm dx} f\left(\frac ab - x\right)
  = f'\left(\frac ab - x\right)
     \frac{\mathrm d}{\mathrm dx} \left(\frac ab - x\right)
  = -f'\left(\frac ab - x\right).
$$
You can continue taking derivatives that way indefinitely, so in general
for positive $k,$
$$ f^{(k)}(x) = \pm  f^{(k)}\left(\frac ab - x\right). $$
In particular this is true when $x = 0$:
$$ f^{(k)}(0) = \pm  f^{(k)}\left(\frac ab\right). $$
We already know all the values on the left are integers,
so the values on the right are integers too.
A: Just take a couple of derivatives-
$$f'(x) = \frac {nx^{n-1}(a-bx)^n - nbx^n(a-bx)^{n-1}}{n!}\\
f''(x) = \frac {n(n-1)x^{n-2}(a-bx)^n - 2nbx^{n-1}(a-bx)^{n-1} + n(n-1)b^2x^n(a-bx)^{n-2}}{n!}$$
Evaluate at $0, \frac {a}{b}$ to see that
$f^{(j)}(0) = 0, f^{(j)}(\frac {a}{b}) = 0$ if $j< n$ and $f^{(j)}(0), f^{(j)}(\frac {a}{b})$ will be an integer in all cases.
Does that help?
