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Can you prove that $e^{n\pi}$ is transcendental $\forall$ algebraic $n \in\mathbb{R}$ $n\neq $ 0 ?

edit : n must be algebraic

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    $\begingroup$ The expression takes all positive real values, so no, clearly it is not transcendental. $\endgroup$ – 5xum Nov 24 '16 at 11:49
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Yes $ e^{\pi n} $ is transcendenatal, for algebraic $ n \neq 0 \in \mathbb{R} $, using the Gelfond–Schneider theorem. This theorem says

If $ a, b $ are algebraic, with $ a \neq 0, 1 $, and $ b $ irrational, then any value of $ a^b $ is transcendental.

Take $ a = e^{i \pi} = -1 $, and $ b = -i n $. Since you assumed that $ n $ is algebraic, we find that $ b = -i n $ is also algebraic. Since $ n \neq 0 $ is real, $ -i n $ must be purely imaginary, so in particular it is irrational. You can then conclude $ a^b = e^{n \pi} $ is transcendental with Gelfond-Schneider.

[In 1900, the special case $ e^{\pi} $ given as an example in Hilbert's 7th problem about whether $ a^b $ is transcendental, if $ a, b $ algebraic, $ a \neq 0, 1 $, $ b $ irrational. Gelfond proved that $ e^{\pi} $ is transcendental in 1929, later seeing this as a special case of the Gelfond-Schendier Theorem of 1934/1935. Without using powerful machinery like the Gelfond-Schneider Theorem, a direct proof that $ e^{\pi n} $ is transcendental is going to be difficult.]

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