Verified answer

e^(iπ) = -1

ln(-1) = iπ

ln(-1) / i = π

ln(-1)(i ⁻¹) = π

Note that i ⁻¹ = -i

since i = √(-1) and 1 / i = 1 / (√(-1)) = 1 / (√(-1))∙((√-1) / (√-1)) since ((√-1) / (√-1)) = 1

1 / (√(-1))∙((√-1) / (√-1)) = 1√(-1) / i² = i / (-1) = -i

∴ (i ⁻¹)ln(-1) = (-i)ln(-1) = π