What's the $n$'th cumulant of the Beta distribution? Is there a closed form for the $n$'th cumulant of the Beta distribution, as a function of the parameters $\alpha$, $\beta$?
The cumulant generating function of the Beta distribution is the logarithm of a confluent hypergeometric function:
$$\log(\mathrm{E}(\mathrm{e}^{tX}))=\log{_1F_1(\alpha,\alpha+\beta ;t)},$$
where $X$ is a random variable following a Beta distribution with parameters $\alpha$, $\beta$ and pdf $f(x)$:
$$f(x)=\frac{x^{\alpha-1}(1-x)^{\beta-1}}{\mathrm{B}(\alpha,\beta)}.$$
So, equivalently, I am asking for a general closed form expression of the coefficient $\kappa_n$ in:
$$\log{_1F_1(\alpha,\alpha+\beta ;t)}=\sum_{n=1}^\infty \kappa_n \frac{t^n}{n!}.$$
Note: By closed form I mean an expression involving no derivatives or integrals, and no infinite sums or products. It can contain finite sums and finite products.
 A: The $n$-th raw moment of the Beta distribution is given by:
$$E[X^n]= \mu_n' = \frac{B(\alpha+n,\beta)}{B(\alpha,\beta)}$$
Thus, to obtain a closed form solution for the $n$-th cumulant, all we need is a method to express the $n$-th cumulant $\kappa_n$ in terms of the raw moments $\mu_i'$ (which we know), and this can be done as follows:
$$\kappa_n = (-1)^{n-1}\left|
\begin{array}{cccc}
 \mu_1' & 1 & 0& 0 & \cdots  & 0 \\
 \mu_2' & \binom{1}{0}\mu_1' & 1& 0 & \cdots  & 0 \\
 \mu_3' & \binom{2}{0}\mu_2' & \binom{2}{1}\mu_1'& 1 & \cdots  & 0 \\
 \vdots  & \vdots  & \vdots & \vdots & & 1 \\
 \mu_n' & \binom{n-1}{0}\mu_{n-1}' & \binom{n-1}{1}\mu_{n-2}'  &\dots &\dots & \binom{n-1}{n-2}\mu_1' \\
\end{array}
\right|$$
This has "no derivatives or integrals, and no infinite sums or products" :)
Example: $n=2$
With $n = 2$, the formula is:
$$\kappa_2 = (-1)^{1}\left|
\begin{array}{cccc}
 \mu_1' & 1  \\
 \mu_2'  &\binom{1}{0}\mu_1' \\
\end{array}
\right|  =  \mu_2' - \mu_1'^2$$
which is the formula for variance, and so must give the correct solution.
Example: $n=3$
With $n = 3$, the formula is:
$$\kappa_3 = (-1)^{2}\left|
\begin{array}{cccc}
 \mu_1' & 1 &0 \\
 \mu_2'  &\binom{1}{0}\mu_1' &1 \\
 \mu_3'  &\binom{2}{0}\mu_2' &\binom{2}{1}\mu_1' \\
\end{array}
\right|  =  2\mu_1'^3 - 3\mu_1' \mu_2' + \mu_3'$$
Automated form
The above was provided because the OP sought a 'closed-form' solution. But, instead of manually deriving each of these cumulant to raw moment conversions, there exist automated ways of doing this that make it much easier. For example, here are the first 5 such conversions using the CumulantToRaw function in mathStatica:

Even with $n = 20$, the conversion takes less than 0.15 seconds to compute.
