Branch Points of the Polylog function

The polylogarithm $$ {\rm Li}_s(z) = \sum_{n=1}^\infty \frac{z^n}{n^s} $$ has obvious branch points at $z=1$.

For integers $s\leq 0$ it is a rational function with a pole of order $1-s$ at $z=1$. If $s=0$ upon integration the function $\frac{1}{1-z}$ turns into a branch point with a cut along the real axis $z \geq 1$.

The principal branch does not contain any other branch points in particular the point $z=0$ is analytic looking at the series. However if one were to go to other sheets, then suddenly the point $z=0$ becomes a second branch point whose cut is typically put along the negative real axis.

Is there a way to clearly see this by looking for a representation valid in another sheet?

For example: If I take ${\rm Li}_s(z)$ for the principal branch, and I start with a path in the upper half plane where the value is supposed to be ${\rm Li}_s(z)$, then the value in the lower half plane after crossing the real axis $z > 1$ is $$ {\rm Li}_s(z) + \frac{2\pi i \ln(z)^{s-1}}{\Gamma(s)} $$ if the function I evaluate this value from is smooth everywhere on the path, in particular on the real axis $z > 1$. From this it is seen that there is a branch point at $z=0$, but how do you get this formula?

The value of the jump can be derived from the integral representation $$\operatorname{Li}_s(z) = \int_0^\infty K_{s, z}(t) dt, \\ K_{s, z}(t) = \frac z {\Gamma(s)} \frac {t^{s - 1}} {e^t - z}.$$ For $z > 1$, $$\operatorname{Li}_s(z \pm i0) = \operatorname{v.\!p.} \int_0^\infty K_{s, z}(t) dt \pm \pi i \operatorname{Res}_{t = \ln z} K_{s, z}(t), \\ \operatorname{Li}_s(z + i0) - \operatorname{Li}_s(z - i0) = \frac {2 \pi i} {\Gamma(s)} \ln^{s - 1} z.$$