Coefficients of $n^{-k}$ in Stirling's Formula I wish to find an explicit formula for $c_1,...,c_k$ such that
\begin{equation}
n!= \sqrt{2\pi n} \Big(\frac{n}{e}\Big)^n\bigg(1+\frac{c_1}{n}+\frac{c_2}{n^2}+...+\frac{c_k}{n^k}+O(n^{-k-1})\bigg).
\end{equation}
This paper gives:
\begin{equation}
a_n=\begin{cases}
    \frac{1}{n+1}a_{n-1}-\sum_{k=2}^{\frac{n}{2}}a_ka_{n+1-k}, & \text{if $n$ even}.\\
    \frac{1}{n+1}a_{n-1}-\sum_{k=2}^{\frac{n-1}{2}}a_ka_{n+1-k}-\frac{1}{2}a_{(n+1)/2}^2, & \text{if $n$ odd}.
  \end{cases}
\end{equation}
Which is then multiplied by a double factorial to give:
\begin{equation}
n!= \sqrt{2\pi n} \Big(\frac{n}{e}\Big)^n\bigg(1+\frac{3a_3}{n}+\frac{3*5*a_5}{n^2}+\frac{3*5*7*a_7}{n^3}+\ldots\bigg).
\end{equation}
Is this the best equation out there at the moment for $n!$? Thanks!

Stirling's formula can be derived applying the Euler-Maclaurin formula to $\ln n! = \sum_{1 \le k \le n} \ln k$, giving an asymptotic formula (the series doesn't converge, but the error is less than the first term left out). The sum is for a given $s$:
$$
\ln n! \approx \ln \sqrt{2 \pi} + (n + 1 / 2) \ln n - n
                 - \sum_{1 \le k \le s}
               \frac{B_{2 k}}{(2 k - 1) 2 k} n^{- 2 k - 1}
$$
Here $B_k$ are the Bernoulli numbers. Take the exponential of this, and expand the exponential of the series as a series...
