$f^{(k)}(x) = \frac{(-1)^{k+1}}{2^k}\cdot\prod^{k-1}_{j=1}(2j-1)\cdot\frac{1}{\sqrt{x^{2k-1}}}$
For $x_0=1$ calculate Taylor series.
So the above expression has already been proven by induction.
$T_{f,x_0}= \sum^\infty_{k=0}\frac{f^{(k)}(x_0)}{k!}\cdot(x-x_0)^k$
$\begin{align} T_{f,1} &= \sum^\infty_{k=0}\frac{f^{(k)}(1)}{k!}\cdot(x-1)^k \\ &= \sum^\infty_{k=0} \frac{(-1)^{k+1}}{2^k}\cdot \prod^{k-1}_{j=1}(2j-1)\cdot \frac{1}{\sqrt{1^{2k-1}}} \cdot\frac{1}{k!}\cdot(x-1)^k \\ &= \sum^\infty_{k=0} \frac{1}{2k-1}\prod^{k}_{j=1}(2j-1) \cdot \frac{1}{k!} \cdot(-1) \cdot \frac{(-1)^k}{2^k} \cdot (x-1)^k \\ &=\sum^\infty_{k=0} \frac{1}{2k-1}\prod^{k}_{j=1}[(2j-1) \cdot \frac{1}{j}] \cdot (-1) \cdot (\frac{(-1)\cdot(x-1)}{2})^k \\ &=\sum^\infty_{k=0} \frac{1}{2k-1}\prod^{k}_{j=1}(2-\frac{1}{j}) \cdot (-1) \cdot (\frac{1-x}{2})^k \end{align}$
Is there a way to transform this expression further
I have the feeling that there is something in particular when suddenly $k!$ disappears.
Starting from the beginning $$f^{(k)}(x) = \frac{(-1)^{k+1}}{2^k}\prod^{k-1}_{j=1}(2j-1)\frac{1}{\sqrt{x^{2k-1}}}=\frac{(-1)^{k+1}}{2^k\sqrt{x^{2k-1}}}\prod^{k-1}_{j=1}(2j-1)$$ Now $$\prod^{k-1}_{j=1}(2j-1)=\frac{2^{k-1} }{\sqrt{\pi }}\Gamma \left(k-\frac{1}{2}\right)$$ makes $$\frac {f^{(k)}(1)}{k!}=\frac{(-1)^{k+1} }{2 \sqrt{\pi }\,k!}\Gamma \left(k-\frac{1}{2}\right)$$ $$T_{f,1}=\frac{1 }{ 2\sqrt{\pi }}\sum_{k=0}^\infty \frac{(-1)^{k+1} }{k!}\Gamma \left(k-\frac{1}{2}\right)(x-1)^k=\sum_{k=0}^\infty \binom{\frac{1}{2}}{k}(x-1)^k$$ which is in fact a very, very simple function.