= Solution
Assume for contradiction that $\pi=a/b$ with positive <integer>[integers] $a,b$. Define
$$
F_n(x)=\frac{b^n}{n!}x^n(\pi-x)^n
=\frac1{n!}x^n(a-bx)^n
$$
and
$$
I_n=\int_0^\pi F_n(x)\sin x\,dx.
$$
The coefficient of $x^{n+j}$ in $F_n$ is
$$
\frac{(-1)^j}{n!}\binom nj a^{n-j}b^j.
$$
Consequently every derivative $F_n^{(k)}(0)$ is an integer: it is zero unless $n\leq k\leq2n$, and in that range the factor $k!/n!$ clears the denominator. Since $F_n(\pi-x)=F_n(x)$, every $F_n^{(k)}(\pi)$ is also an integer.
Repeated <integration by parts>, ending when derivatives above degree $2n$ vanish, gives
$$
I_n=\sum_{j=0}^n(-1)^j
\left(F_n^{(2j)}(0)+F_n^{(2j)}(\pi)\right).
$$
Thus $I_n$ is an integer. It is strictly positive because $F_n(x)\sin x>0$ for $0<x<\pi$.
On the other hand, $x(\pi-x)\leq\pi^2/4$, so
$$
0<I_n
\leq\frac{\pi}{n!}\left(\frac{b\pi^2}{4}\right)^n.
$$
The <factorial-over-power series> implies that the right-hand side tends to zero. For all sufficiently large $n$ this says that the positive integer $I_n$ is less than one, a contradiction. Therefore the <Irrationality of pi> is established:
$$
\boxed{\pi\notin\mathbb Q.}
$$
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