Problem: Prove that $ 2 = 4$. Solution: Consider the value of the following infinitely iterated exponent: $$\displaystyle \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\cdot^{\cdot^{\cdot}}}}}$$ Let $ a_n = \sqrt{2} \uparrow \uparrow n$, that is, the above power tower where we stop at the $ n$-th term. Then $ a_n$ is clearly an increasing sequence, and moreover $ a_n \leq 4$ by a trivial induction argument: $ \sqrt{2} \leq 4$ and if $ a_n \leq 4$ then $ a_{n+1} = (\sqrt{2})^{a_n} \leq (\sqrt{2})^{4} = 4$....