(1)

\begin{align*} \Li_{n}\left(-z\right) & =\sum_{k=1}^{\infty}\frac{\left(-z\right)^{k}}{k^{n}}\\ & =\sum_{k=1}^{\infty}\left(\frac{\left(-z\right)^{2k-1}}{\left(2k-1\right)^{n}}+\frac{\left(-z\right)^{2k}}{\left(2k\right)^{n}}\right)\\ & =\sum_{k=1}^{\infty}\left(\frac{\left(-z\right)^{2k-1}}{\left(2k-1\right)^{n}}-\frac{\left(-z\right)^{2k}}{\left(2k\right)^{n}}+2\frac{\left(-z\right)^{2k}}{\left(2k\right)^{n}}\right)\\ & =\sum_{k=1}^{\infty}\left(\frac{-z^{2k-1}}{\left(2k-1\right)^{n}}+\frac{-z^{2k}}{\left(2k\right)^{n}}+2\frac{\left(z^{2}\right)^{k}}{\left(2k\right)^{n}}\right)\\ & =\sum_{k=1}^{\infty}\left(\frac{-z^{k}}{k^{n}}+\frac{\left(z^{2}\right)^{k}}{2^{n-1}k^{n}}\right)\\ & =-\Li_{n}\left(z\right)+\frac{\Li_{n}\left(z^{2}\right)}{2^{n-1}} \end{align*} となるので、
\[ \Li_{n}\left(z\right)+\Li_{n}\left(-z\right)=\frac{1}{2^{n-1}}\Li_{n}\left(z^{2}\right) \] となり与式は成り立つ。

(2)

\begin{align*} \Log\left(1-\frac{1}{z}\right) & =\Log\left(\frac{z-1}{z}\right)\\ & =\left[\Log\left(\frac{z-1}{z}\right)\right]_{\infty}^{z}\\ & =\int_{\infty}^{z}\left(\frac{d}{dz}\Log\left(\frac{z-1}{z}\right)\right)dz\\ & =\int_{\infty}^{z}\left(\left(\frac{z}{z-1}\right)\frac{z-\left(z-1\right)}{z^{2}}\right)dz\\ & =\int_{\infty}^{z}\left(\frac{1}{z\left(z-1\right)}\right)dz\\ & =\int_{\infty}^{z}\left(\frac{1}{z-1}-\frac{1}{z}\right)dz\\ & =\left[\Log\left(z-1\right)-\Log\left(z\right)\right]_{\infty}^{z}\\ & =\Log\left(z-1\right)-\Log\left(z\right) \end{align*}

(3)

\begin{align*} \Li_{2}\left(1-z\right) & =\int_{0}^{1-z}\frac{\Li_{1}\left(z\right)}{z}dz\\ & =-\int_{0}^{1-z}\frac{\Log\left(1-z\right)}{z}dz\\ & =-\left[\Log\left(z\right)\Log\left(1-z\right)\right]_{0}^{1-z}-\int_{0}^{1-z}\frac{\Log\left(z\right)}{1-z}dz\\ & =\Li_{1}\left(z\right)\Log\left(z\right)+\int_{1}^{z}\frac{\Log\left(1-z\right)}{z}dz\\ & =\Li_{1}\left(z\right)\Log\left(z\right)-\int_{1}^{z}\frac{\Li_{1}\left(z\right)}{z}dz\\ & =\Li_{1}\left(z\right)\Log\left(z\right)-\Li_{2}\left(z\right)+\Li_{2}\left(1\right)\\ & =\Li_{1}\left(z\right)\Log\left(z\right)-\Li_{2}\left(z\right)+\zeta\left(2\right)\\ & =\frac{\pi^{2}}{6}+\Log\left(z\right)\Li_{1}\left(z\right)-\Li_{2}\left(z\right) \end{align*}

(4)

\begin{align*} \Li_{2}\left(1-z\right)+\Li_{2}\left(1-\frac{1}{z}\right) & =\frac{\pi^{2}}{6}+\Log\left(z\right)\Li_{1}\left(z\right)-\Li_{2}\left(z\right)+\frac{\pi^{2}}{6}+\Log\left(\frac{1}{z}\right)\Li_{1}\left(\frac{1}{z}\right)-\Li_{2}\left(\frac{1}{z}\right)\\ & =\frac{\pi^{2}}{6}+\Log\left(z\right)\Li_{1}\left(z\right)-\Li_{2}\left(z\right)+\frac{\pi^{2}}{6}+\Log\left(\frac{1}{z}\right)\Li_{1}\left(\frac{1}{z}\right)+\Log\left(z\right)\Li_{1}\left(\frac{1}{z}\right)-\Log\left(z\right)\Li_{1}\left(\frac{1}{z}\right)-\Li_{2}\left(\frac{1}{z}\right)\\ & =\frac{\pi^{2}}{3}+\Log\left(z\right)\left(\Li_{1}\left(z\right)-\Li_{1}\left(\frac{1}{z}\right)\right)-\left(\Li_{2}\left(z\right)+\Li_{2}\left(\frac{1}{z}\right)\right)+\left(\Log\left(z\right)+\Log\left(\frac{1}{z}\right)\right)\Li_{1}\left(\frac{1}{z}\right)\\ & =\frac{\pi^{2}}{3}+\Log\left(z\right)\left(-\log\left(1-z\right)+\log\left(1-\frac{1}{z}\right)\right)-\left(2\Li_{2}\left(1\right)-\frac{\Log^{2}\left(z\right)}{2}-\Log\left(z\right)\left(\Log\left(1-z\right)-\Log\left(z-1\right)\right)\right)+\left(\Log\left(z\right)+\Log\left(\frac{1}{z}\right)\right)\Li_{1}\left(\frac{1}{z}\right)\\ & =\frac{\pi^{2}}{3}+\Log\left(z\right)\left(-\log\left(1-z\right)+\log\left(z-1\right)-\log\left(z\right)\right)-2\Li_{2}\left(1\right)+\frac{\Log^{2}\left(z\right)}{2}+\Log\left(z\right)\left(\Log\left(1-z\right)-\Log\left(z-1\right)\right)+\left(\Log\left(z\right)+\Log\left(\frac{1}{z}\right)\right)\Li_{1}\left(\frac{1}{z}\right)\\ & =-\frac{1}{2}\Log^{2}\left(z\right)+\left(\Log\left(z\right)+\Log\left(\frac{1}{z}\right)\right)\Li_{1}\left(\frac{1}{z}\right) \end{align*}