$\pi$ is not algebraic of degree one or two

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Can someone help me understand the explanations made in the last two paragrapghs, starting from the sentence "Since the factors $P(x)$ and $P(x)-P(0)...$"

I think I need to be more spesific to get an answer..
First of all, can you show that the expressions $P(x)$ and $P(x)-P(0)$ in $f^{(2k)}(x)$ are nonzero for $x=0$ and $x=\pi$? And they must have a factor of $(2n)!$ in the numerator?