Help needed !!!.

Can this identity be proven, and if so how?  π^2/6 = 1/6 (-i log(-1))^2

\(\begin{array}{|rcll|} \hline && \frac{1}{6}\cdot \Big(-i \log(-1) \Big)^2 \\ && &\small{ \begin{array}{|rcll|} \hline \log(z) &=& \ln|z| + i\underbrace{\arg(z)}_{=\arctan(\frac{b}{a}) } \quad | \quad z = a+b\cdot i \\ \log(-1) &=& \ln|-1| + i\underbrace{\arg(-1)}_{= \underbrace{\arctan\left(\frac{0}{-1}\right)}_{=-\pi} } \quad | \quad z = -1+0\cdot i \\ \log(-1) &=& \ln|-1| + i\cdot(-\pi) \\ \log(-1) &=& \ln(1) - i\cdot \pi \quad | \quad \ln(1) = 0 \\ \log(-1) &=& 0 - i\cdot \pi \\ \log(-1) &=& - i\cdot \pi \\ \hline \end{array}} \\ &=& \frac{1}{6}\cdot \Big(-i \cdot(- i\cdot \pi) \Big)^2 \\ &=& \frac{1}{6}\cdot (i^2 \cdot \pi )^2 \quad | \quad i^2 = -1 \\ &=& \frac{1}{6}\cdot (-\pi )^2 \\ &=& \frac{1}{6}\cdot \pi ^2 \\ &=& \dfrac{\pi ^2}{6} \\ \hline \end{array}\)

Brilliant heureka, as usual. Thank you very much.

\(\because e^{i\pi} = -1\\ \text{We can immediately imply that} \log(-1) = i\pi\\ \dfrac{1}{6}(-i\log(-1))^2 \\ =\dfrac{1}{6}(-i\cdot i\pi)^2\quad\boxed{-i\cdot i = 1}\\ =\dfrac{1}{6}\pi^2\\ \)

Hence, proved.

Max: You are simply beyond words !! Bravo and thanks.