# Is $e^{n\pi}$ transcendental?

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

edit : n must be algebraic

• The expression takes all positive real values, so no, clearly it is not transcendental. – 5xum Nov 24 '16 at 11:49

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.]