I have this problem where I need to rewrite the logarithmic form of Stirling's series
$$\ln(z!)=\frac{1}{2}\ln(2\pi)+(z+\frac{1}{2})\ln(z)-z+\frac{1}{12z}-\frac{1}{360z^3}+\frac{1}{1260z^5}-...$$
First I thought to use the logarithm properties:
$$\ln(z!)=\ln(\sqrt{2\pi}z^{(z+\frac{1}{2})})-\ln(e^z)+\frac{1}{12\ln(e^z)}-\frac{1}{360z^2\ln(e^z)}+\frac{1}{1260z^4\ln(e^z)}-...$$
$$\Rightarrow \ln(z!)=\ln(\sqrt{2\pi}z^{(z+\frac{1}{2})})+\ln(e^{-z})+\frac{\ln(e^{-z})}{12}-\frac{\ln(e^{-z})}{360z^2}+\frac{\ln(e^{-z})}{1260z^4}-...$$
$$\Rightarrow \ln(z!)=\ln(\sqrt{2\pi}z^{(z+\frac{1}{2})})+\ln(e^{-z})(1+\frac{1}{12}-\frac{1}{360z^2}+\frac{1}{1260z^4}-...$$
$$\Rightarrow \ln(z!)=\ln(\sqrt{2\pi}z^{(z+\frac{1}{2})}e^{-z})(1+\frac{1}{12}-\frac{1}{360z^2}+\frac{1}{1260z^4}-...$$
I get something similar but not quite right. The term of $\sqrt{2\pi}z^{(z+\frac{1}{2})}e^{-z}$ is really similar to the Stirling formula $n!\approx \sqrt{2\pi n}*n^{(n)}e^{-n}$ but I still cannot get an idea to get the terms inside the parentheses of the answer the problem is giving me.
We have $$ \log (z!) \sim \tfrac{1}{2}\log (2\pi ) + \left( {z + \tfrac{1}{2}} \right)\log z - z + \frac{1}{{12z}} - \frac{1}{{360z^3 }} + \ldots . $$ Taking the exponential of both sides gives \begin{align*} z! & \sim \exp \left( {\tfrac{1}{2}\log (2\pi ) + \left( {z + \tfrac{1}{2}} \right)\log z - z + \frac{1}{{12z}} - \frac{1}{{360z^3 }} + \ldots } \right) \\ & = \exp \left( {\log \left( {\sqrt {2\pi } z^{z + 1/2} \mathrm{e}^{ - z} } \right)} \right)\exp \left( {\frac{1}{{12z}} - \frac{1}{{360z^3 }} + \ldots } \right) \\ & = \sqrt {2\pi } z^{z + 1/2} \mathrm{e}^{ - z} \exp \left( {\frac{1}{{12z}} - \frac{1}{{360z^3 }} + \ldots } \right). \end{align*} To finish the derivation, we use the Maclaurin series of the exponential: \begin{align*} \exp \left( {\frac{1}{{12z}} - \frac{1}{{360z^3 }} + \ldots } \right) & = 1 + \left( {\frac{1}{{12z}} - \frac{1}{{360z^3 }} + \ldots } \right) + \frac{{\left( {\frac{1}{{12z}} - \frac{1}{{360z^3 }} + \ldots } \right)^2 }}{2} + \ldots \\ & = 1 + \frac{1}{{12z}} + \frac{1}{{288z^2 }} - \frac{{139}}{{51840z^3 }} - \ldots . \end{align*}