Pi is Transcendental

Theorem
$\pi$ (pi) is transcendental.

Proof
Proof by Contradiction:

$\pi$ is not transcendental.

Hence by definition, $\pi$ is algebraic.

Let $\pi$ be the root of a non-zero polynomial with rational coefficients, namely $f \left({x}\right)$.

Then, $g \left({x}\right) := f \left({ix}\right) f \left({-ix}\right)$ is also a non-zero polynomial with rational coefficients such that:
 * $g \left({i \pi}\right) = 0$

Hence, $i \pi$ is also algebraic.

From the Weaker Hermite-Lindemann-Weierstrass Theorem, $e^{i \pi}$ is transcendental.

However, from Euler's Identity:
 * $e^{i \pi} = -1$

which is the root of $h \left({z}\right) = z + 1$, and so is algebraic.

This contradicts the conclusion that $e^{i \pi}$ is transcendental.

Hence by Proof by Contradiction it must follow that $\pi$ is transcendental.