For $U$ a unitary $N \times N$ matrix, randomly distributed according to Haar measure, we have the complex-valued random variable $\log (\det (1-U))$. The real part and imaginary parts of $\log (\det (1-U))$ are real-valued random variables. (For the imaginary part, it is necessary to choose a branch of the logarithm. Integrating $\frac{d}{dt} \log (\det (1-t U))$ for $t$ from $0$ to $1$ provides a good choice.)

Keating and Snaith in Random matrix theory and $\zeta(1/2+it)$ exactly calculated the cumulants of the real and imaginary parts of $\log(\det (1-U))$, giving exact formulas and obtaining large $N$ limits. In particular, for the $n$th cumulant of the real part, $n>2$, they obtained a value of

$$(-1)^n \frac{2^{n-1}-1} {2^{n-1} } \zeta(n-1) \Gamma(n) + O (N^{2-n})$$

and for the $n$th cumulant of the real part, $n>2$ even, they obtained a value of $$\frac{ (-1)^{\frac{n}{2}+ 1}}{ 2^{n-1}} \zeta(n-1) \Gamma(n) + O (N^{2-n}).$$

Their proof used the Selberg integral, and had applications to the study of the Riemann zeta function.

For $\lambda_1,\dots, \lambda_N$ the eigenvalues of $U$, we of course have $\log (\det(1-U))= \sum_{i=1}^N \log (1-\lambda_i)$. Using this, we can express the $n$th cumulant as an expectation as a sum over $k$-tuples of eigenvalues for $k\leq n$. Since the eigenvalues of a Haar-random unitary matrix are a determinantal point process, any such expectation can be expressed as an integral against determinants of $k\times k$ matrices, $k\leq n$, with entries given by the kernel $K_N (\theta, \theta') = \sum_{j=0}^{N-1} e^{ i j (\theta-\theta')}$. In fact, in the case of cumulants specifically, this expression simplifies somewhat, and I believe you end up with just the terms in the Leibniz formula arising from cyclic permutations.

My question is:

Does there exist an alternate proof of the Keating-Snaith formulas (maybe with a less precise error term) via the determinantal point process description of the eigenvalues of random matrices?

It is straigtforward to get the second moment/cumulant formula this way, so the question is primarily about the higher cumulants.