My favorite proof, x=0 for all x

This proof uses the imaginary number $ i = \sqrt{-1} $

Let's start with something normal
\[ x = x \]
Then exponentiation of both sides:
\[ e^x = e^x \]
Multiply the right side 'x' by 1 with $ 1 = \frac{2 \pi i}{2 \pi i} $
\[ e^x = e^{( \frac{2 \pi i}{2 \pi i} x )} \]
Factor out the $ 2 \pi i $
\[ e^x = ( e^{2 \pi i} )^{( \frac{x}{2 \pi i} )} \]
Then use Euler's Identity $ e^{2 \pi i} = 1 $
\[ e^x = 1^{(\frac{x}{2 \pi i} )} \]
But 1 to any power is 1 so
\[ e^x = 1 \]
Then take the logarithm of both sides gives
\[ x = 0 \; \forall \: x \]
Or x = 0 for all x