Pi is Transcendental

Theorem
$\pi$ (pi) is transcendental.

Proof
Assume the contrary.

Since $i$ is a root of $z^2 + 1 = 0$, it is algebraic.

Then, from Algebraic Numbers are Closed under Multiplication, $i \pi$ is also algebraic.

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

However, from Euler's Identity, $e^{i \pi} = -1$ which is the root of $z + 1 = 0$, which is algebraic.

Hence, there is a contradiction.