Proof that \(\pi\) is irrational.

Assume \(\pi\) is rational, that is, assume it is of the form \(\frac{a}{b}\) where \(a\) and \(b\) are both positive integers. Let

\[\begin{align} f(x) &= \frac{x^n (a-bx)^n}{n!} \\ F(x) &= f(x) + \cdots + (-1)^j f^{[2j]}(x) + \cdots + (-1)^n f^{[2n]}(x) \end{align}\]

where \(f^{[k]}\) denotes \(k\)-th derivative of \(f\).

\(f(x)\) has integer coefficients except \(\frac{1}{n!}\)
\(f(x) = f(\pi - x)\)
\(0 \leq f(x) \leq \frac{\pi^n a^n}{n!}\) for \(0 \leq x \leq \pi\)
For \(0 \leq j < n\), the \(j\)-th derivative of \(f\) equals 0 at 0 and \(\pi\)
For \(j \geq n\), the \(j\)-th derivative of \(f\) is integer at 0 and \(\pi\) (from 1. above)
\(F(0)\), \(F(\pi)\) is integer (from 4., 5. above)
\(F(x) + F''(x) = f(x)\)
\((F'(x) \sin x - F(x) \cos x)' = f(x) \sin x\) (from 7. above)
\( \int_0^\pi f(x) \sin x\) is an integer
For large \(n\), this integral is between 0 and 1 (from 3. above)

Contradiction. So \(\pi\) is irrational.