Is this proof about $\pi$ is irrational correct? A proof is
If $\pi$ is rational, then $\pi$=$\frac{a}{b}$ where $a,b \in \mathbb{N}$ 
let $f(x)=x^n\left[\dfrac{(a-bx)^n}{n!}\right]$ , for $0<x<\frac{a}{b}$
$0<f(x)<\dfrac {\pi^na^n} {n!}$ , $0<\sin x<1$
thus $0<f(x)\sin x<\dfrac {\pi^na^n} {n!}$
where $n$ is big enough 
$0<\int _{0}^{\pi}f(x)\sin x\mbox{d}x<1 $    $\left( 1\right)$
let $F(x)=\sum _{i=0}^{n}(-1)^if^{(2i)}\left
( x\right)$ ($f^{(i)}\left( x\right)$ means $i$th derivative )
because $n!f(x)$ is a polynomial with integer coefficient, and all the degrees  is bigger than n,
so $f(x)$ and its i th derivatives at $x-0$ are all integer 
thus $F(0)$ and $F(\pi)$ are also integers
$\dfrac{\mbox{d}}{\mbox{d}x}[F'(x)\sin x-F(x)\cos x]=F''(x)\sin x+F'(x)\cos x-F'(x)\cos x+F(x)\sin x=F''(x)\sin x+F(x)\sin x=f(x)\sin x$
thus $\int _{0}^{\pi}f(x)\sin x\mbox{d}x=\int_{0}^{\pi}[F'(x)\sin x-F(x)\cos x]\mbox{d}x==F(π)+F(0) $
so $\int _{0}^{\pi}f(x)\sin x\mbox{d}x$ is integer  contradicts  with $(1)$.
I think this proof has some error but I cannot find it. 

This proof is due to Niven. See here.
