Theorem. $\pi$ is irrational.
Assume that $\pi$ is rational. Then $\pi = q/p$ for some positive integer $p, q$. Let $n\in\mathbb{N}$ and define a polynomial $f_n(x)$ by
$$f_n(x) = \frac{p^{n}x^{n}(\pi-x)^{n}}{n!} = \frac{x^{n}(q-px)^{n}}{n!}.$$
Since the polynomial $x(q-px)$ has a maximum value at the point $x = \frac{q}{2p} = \frac{\pi}{2}$, $f_n(x) \leq f_n\left(\frac{\pi}{2}\right)$ for any $0\leq x\leq \pi$. Let $M_n$ be the maximum value of $f_n(x)$. Then
$$M_n = f_n\left(\frac{\pi}{2}\right) = \frac{p^{n}\pi^{2n}}{n!2^{2n}}.$$
Since
$$\lim_{n\rightarrow\infty}\frac{a^n}{n!} = 0$$
for any positive number $a$, we obtain $\lim_{n\rightarrow\infty}M_n = 0$.
Define $I_n$ by
$$I_n = \int_0^\pi f_n(x)\sin xdx.$$
Since $f_n(x)>0$ and $\sin x > 0$ for any $x\in(0,\pi)$, we have $I_n>0$. Since $f_n(x)\leq M_n$ for any $n\in\mathbb{N}$ and $0\leq x\pi$,
$$I_n \leq \int_0^\pi M_n\sin xdx = 2M_n.$$
Thus $0<I_n\leq 2M_n$ for any $n\in\mathbb{N}$. Since $M_n\rightarrow 0$ as $n\rightarrow\infty$, there is $N\in\mathbb{N}$ such that $M_n < 1/4$ for all $n\geq N$. Thus $0<I_n<1$ for any $n\geq N$.
On the other hand, Note that $f^{(k)}(x) = 0$ if $k > 2n$. Then
$$I_n = \int_0^\pi f_n(x)\sin xdx = \left[-f_n(x)\cos x + f'_n(x)\sin x - \cdots \pm f_n^{(2n)}(x)\cos x\right]_0^\pi.$$
Note that $f^{(k)}(0) = f^{(k)}(\pi) = 0$ if $0\leq k< n$, by the definition of the polynomial $f_n(x)$. If $n\leq k< 2n-1$, then
$$\begin{eqnarray}
f_n^{(k)}(x) & = & p^{n}n(n-1)\cdots(n-(k-n))\left[(-1)^{k-n}(\pi-x)^{n-(k-n)} + (-1)^nx^{n-(k-n)}\right] \\
& & +\sum_{i=k-(n-1)}^{n-1}c_ix^{n-i}(\pi-x)^{n-(k-i)}
\end{eqnarray}$$
for some numbers $c_i$. Thus
$$\begin{eqnarray}
f_n^{(k)}(0) & = & n(n-1)\cdots(2n-k)(-1)^{k-n}p^{k-n}q^{2n-k} \\
f_n^{(k)}(\pi) & = & n(n-1)\cdots(2n-k)(-1)^np^{k-n}q^{2n-k}
\end{eqnarray}$$
if $n\leq k<n-1$. Note that
$$f_n^{(2n-1)}(x) = p^nn!\left[(-1)^{n-1}(\pi-x) - x\right]$$
and so $f_n^{(2n-1)}(0) = n!(-1)^{n-1}p^{n-1}q$ and $f_n^{(2n-1)}(\pi) = n!p^{n-1}q^n$. Note also that
$$f^{(2n)}_n(x) = (-1)^np^{n}n!.$$
Thus $f^{(k)}(\pi)$ and $f^{(k)}(0)$ are integer for any $0\leq k\leq 2n$. Since
$$\begin{eqnarray}
I_n & = & \left[-f_n(x)\cos x + f'_n(x)\sin x - \cdots \pm f_n^{(2n)}(x)\cos x\right]_0^\pi \\
& = & \sum_{n\leq k\leq 2n, \ \textrm{$k$ is even}}(-1)^{k/2+1}\left[f_n^{(k)}(\pi)\cos\pi - f_n^{(k)}(0)\cos 0\right] \\
& = & \sum_{n\leq k\leq 2n, \ \textrm{$k$ is even}}(-1)^{k/2}\left[f_n^{(k)}(0) + f_n^{(k)}(\pi)\right],
\end{eqnarray}$$
$I_n$ must be an integer. But it contradicts that $0<I_n<1$ if $n\geq N$. $\Box$
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