Exercise 5, § 2

Show that $\pi$ is irrational, in the following way: if one had $\pi = p/q$ ($p$ and $q$ integers), then on putting $f(x) = (x(\pi - x))^n / n!$, the integral $q^n \int_0^\pi f(x) \sin x \, dx$ would be an integer $> 0$ (use the formula for integration by parts of order $n + 1$); but show that on the other hand $q^n \int_0^\pi f(x) \sin x \, dx$ tends to 0 as $n$ tends to $+\infty$.