\begin{equation} \label{firseq} 0<f(x)<\frac{1}{n!} \end{equation} For a contradiction, suppose $\pi^2$ is rational, so that $\pi^2=\frac{a}{b}$ , where $a,b$ are positive integers.

For $x\in (0,1)$ let us define $$G(x)=b^n[\pi^{2n}f(x)-\pi^{2n-2}f''(x)+\pi^{2n-4}f^{(4)}(x)-...+(-1)^nf^{(2n)}(x)].$$ We have that $f(0)=0$ and $f^{(m)}(0)=0$ if $m<n$ or $m>2n$ . But, if $n \leq m \leq 2n$ , then $$f^{(m)}(0)=\frac{m!}{n!}c_m,$$ an integer. Hence $f(x)$ and all its derivates take integral values at $x=0$ .Since $f(1-x)=f(x)$ , the same is true at $x=1$

so that $G(0)$ and $G(1)$ are integers. We have

\begin{eqnarray*} \frac{d}{dx}[G'(x)\sin{\pi x}-\pi G(x)\cos{\pi x}] &=& [G''(x)+\pi^2G(x)]\sin{\pi x} \\ &=& b^n\pi^{2n+2}f(x)\sin{\pi x} \\ &=& \pi^2a^n \sin{\pi x}f(x). \end{eqnarray*} Hence $$\pi\int_0^1a^n \sin{\pi x}f(x)dx=[\frac{G'(x)\sin{\pi x}}{\pi}-G(x)\cos{\pi x}]_0^1$$ $$=G(0)+G(1),$$ witch is an integer. But by equation , $$0<\pi\int_0^1a^n \sin{\pi x}f(x)dx<\frac{\pi a^n}{n!}<1.$$ For a large enough $n$ , we obtain a contradiction.

For any integer $n$ , if $a^n$ is irrational then a is irrational (proof), and since $\pi^2$ is irrational $\sqrt{\pi^2}=\pi$ is also irrational.