(1)

\begin{align*} \int\Li_{n}\left(e^{z}\right)dz & =\int\frac{\Li_{n}\left(\xi\right)}{\xi}d\xi\cmt{e^{z}=\xi}\\ & =\Li_{n+1}\left(\xi\right)+C\\ & =\Li_{n+1}\left(e^{z}\right)+C \end{align*}

(2)

(1)より、
\begin{align*} \int\Li_{n}\left(\alpha e^{\beta z}\right)dz & =\frac{1}{\beta}\int\Li_{n}\left(e^{\xi}\right)d\xi\cmt{\alpha e^{\beta z}=e^{\xi}}\\ & =\frac{1}{\beta}\Li_{n+1}\left(e^{\xi}\right)+C\\ & =\frac{1}{\beta}\Li_{n+1}\left(\alpha e^{\beta z}\right)+C \end{align*} となるので与式は成り立つ。