The series expansion of $\frac{1}{\sqrt{e^{x}-1}}$ at $x=0$

The function $ \displaystyle\frac{1}{\sqrt{e^{x}-1}}$ doesn't have a Laurent expansion at $x=0$.

But according to Wolfram Alpha, it does have a series expansion that includes terms raised to noninteger powers. Specifically, $\displaystyle\frac{1}{\sqrt{x}}- \frac{\sqrt{x}}{4} + O(x^{\frac{3}{2}})$.

How is that series derived?

My initial thought was to use general binomial theorem. But I don't seem to get anywhere with that.

Related problems: (I), (II). Just find the Taylor series of the function

$$ \frac{\sqrt{x}}{\sqrt{e^{x}-1}}. $$

Added: Here is a formula for the $n$th derivative of the function

$$\left(\frac{\sqrt{x}}{\sqrt{e^{x}-1}}\right)^{(n)}= \frac{\pi}{2}\sum _{k=0}^{n} \sum _{i=0}^{k} \sum _{m=0}^{i}{ \frac { \binom {k}{i}\left[\matrix{i\\m}\right] \left\{\matrix{n\\k}\right\} x^{\frac{1}{2}-m}\, {\rm e}^{(k-i) x} \left( {\rm e}^x - 1 \right)^{i-k-\frac{1}{2}} }{\Gamma \left( \frac{1}{ 2}-k+i \right) \Gamma \left( \frac{3}{2}-m \right) }} .$$

Note: I'll appreciate it if someone can verify this formula with Maple or Mathematica.

Binomial expansion version... as $x \to 0^+$, $$ e^x-1 = \left(1 + x +\frac{x^2}{2} + \frac{x^3}{6} +\dots\right) - 1 = x \left(1+\frac{x}{2}+\frac{x^2}{6}+\dots\right) = x(1+S), $$ where $S \to 0$. So $$ \left(e^x-1\right)^{-1/2} = x^{-1/2}\left(1+S\right)^{-1/2} = x^{-1/2}\sum_{k=0}^\infty \binom{-1/2}{k} S^k $$ Then put in what $S$ is (as many terms as needed...)