[จูกัดเหลียง]

Let $\pi = \dfrac{a}{b}$ where $a,b$ is the natural number with $\gcd(a,b)=1$
Consider the polynomial $f(x)=\left[\,\dfrac{x^n(a-bx)^n}{n!}\right] $ it can be seen that $f(x)=f(\pi-x)$
$\Longrightarrow f^{(k)}(0)\in\mathbb{Z}$ for any natural number $k$ , Moreover , $f^{(k)}(\pi-x)=(-1)^kf^{(k)}(x)$, Consequently, $f^{(k)}(\pi)$ is also the integer number.

Let us consider the function $\displaystyle g(x)=\sum_{k=0}^n \Big( (-1)^kf^{(2k)}(x)\Big)$ We have that, $\displaystyle g(x)+g^{(2)}(x)=f(x)$
In addition, $f(x)\sin x=\Big(g^{'}(x)\sin x-g(x)\cos x\Big)^{'}$ Or equivalently, $\displaystyle p=\int_{0}^{\pi} \Big(f(x)\sin x\Big) dx=g(\pi)+g(0) \in\mathbb{Z}$
However, $\displaystyle 0<\int_{0}^\pi f(x)\sin x dx=\int_{0}^\pi \left[\,\dfrac{x^n(a-bx)^n}{n!}\right]\sin(x) dx \le \int_{0}^{\pi}\frac{(a\pi)^n}{n!} dx=\frac{(a\pi)^n}{n!}\pi$
Because of the fact that, $\displaystyle \lim_{n\rightarrow \infty}\dfrac{(a\pi)^n}{n!}=0$

Hence, for some great enough integer $n$, $0<p<1$ which is contradiction, Hence, $\pi$ is irrational.