a simple proof that $\pi$ is irrational by Ivan Niven The following is Ivan Niven's simple proof that $\pi$ is rational:

Here I didn't understand this part:
For $0\lt x\lt \pi,$  $$0\lt f(x)\sin x\lt {\pi^na^n\over n!}$$
First of all how he concluded this inequality?
Secondly knowing that for sufficiently large n, the integral in (1) is arbitrarily small ; how he concluded that (1) is false and thus $\pi$ is irrational?
 A: For $0 < x < π$, note that $\displaystyle π = \frac{a}{b}$, \begin{align*}0 < f(x) \sin x \leqslant f(x) &= \frac{1}{n!} x^n (a - bx)^n = \frac{1}{n!} \left(-b\left(x - \frac{a}{2b}\right)^2 + \frac{a^2}{4b}\right)^n\\
&\leqslant \frac{1}{n!} \left(\frac{a^2}{4b}\right)^n = \frac{1}{n!} \left(\frac{πa}{4}\right)^n < \frac{π^n a^n}{n!}.
\end{align*}
For the second question, a positive integer must be at least $1$, but the integral tends to $0$ when $n \to \infty$, which is a contradiction.
A: First regarding the inequality : 
Consider the polynomial $f(x)$ on the interval $[0,\pi]$, 
$$ f(x) = \frac{x^n(a-bx)^n}{n!}, $$
$$ f(x) = \frac{x^n a^n (1-bx/a)^n}{n!}, $$
$$ f(x) = \frac{x^n a^n (1-x/\pi)^n}{n!}, $$
Now note that $0\leq x/\pi \leq 1$ which implies that $0 \leq (1-x/\pi) \leq 1$
$$ f(x) = \frac{x^n a^n }{n!} (1-x/\pi)^n \leq \frac{x^n a^n }{n!} , $$
Now just note that $x<\pi$.
$$ f(x) = \frac{x^n a^n }{n!} (1-x/\pi)^n \leq \frac{\pi^n a^n }{n!} , $$

Regarding the conclusion : 
Niven showed three properties of the integral which are incompatible. 
1) The integral is positive for all $n$. 
2) The integral is an integer for all $n$. 
3) The integral can be made arbitrarily small for sufficiently large $n$.
The smallest positive integer is $1$. Conclusion (3) tells us that there is a large enough $n$ to make the integral smaller than $1$. Therefore conclusion (2) and conclusion (3) contradict each other. 
This contradiction is a consequence of the assumption that $\pi$ is rational, so we conclude that $\pi$ is not rational. 
