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How do you prove $\pi^5$ is transcendental? You may assume $\pi$ is transcendental

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closed as off-topic by user99914, Watson, Qwerty, Andrew D. Hwang, user91500 Nov 26 '16 at 13:59

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    $\begingroup$ Please do not vandalize your posts. $\endgroup$ – Pierre-Guy Plamondon Nov 26 '16 at 12:19
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If $\pi^5$ were algebraic, then it would be a zero of a polynomial with rational coefficients $$P(x)=a_nx^n+\dots+a_1x+a_0.$$ Then, $\pi$ would be algebraic as a zero of $P(x^5)=\sum_{k=0}^n a_kx^{5k}$. Absurd.

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Otherwise there's a non-zero rational polynomial $\;p(x)\;$ of which it is a root, but then

$$p(x)=\sum_{n=0}^m a_nx^n\implies 0=p(\pi^5)=\sum_{n=0}^ma_n \pi^{5n}\implies \pi \;\;\text{ is algebraic...}$$

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