Is the Pattern in the Number of Digits in the Bernoulli Numbers Showing Something Significant For the first couple of powers of $10$, the number of digits in these show a certain pattern, is this a coincidence or is their a reasonable explanation. Specifically if we look at
$$ \lfloor \log_{10}\left|B\left(10^n\right)\right| \rfloor $$
Where $ B(k) $ is the kth Bernoulli number and n is a positive integer.
For values of n between $2$ and $8$, the number of digits are as follows:
$$ 078,\ 1769,\ 27677,\ 376755,\ 4767529,\ 57674260,\ 676752569 $$
For each of the number of digits, the pattern seems to be start with a leading digit of $ n-2$ and then carry on a pattern of adding a digit which the next element will repeat, and then add a new one. This leads to a number that seems to begin $76752(5/6)6...$ For $n=7$ this seems to break up a bit and give a digit of $6$ when $n=8$ implies that it should be $5$. MY question is three parted; does this pattern continue, if so, what number is it creating as $n \to\infty$, and is there any reason for why this pattern is what it is?

The reason is the asymptotic equality(1)
$$\lvert B(2k)\rvert \sim \frac{2\cdot(2k)!}{(2\pi)^{2k}}.$$
Thus we obtain
$$\log_{10} \lvert B(2k)\rvert = \log_{10} \left((2k)!\right) + \log_{10} 2 - 2k\log_{10}(2\pi) + o(1),$$
and using Stirling's approximation
$$\log (n!) = \left(n+\tfrac{1}{2}\right)\log n - n + \tfrac{1}{2}\log (2\pi) + O(n^{-1})$$
that becomes
$$\begin{aligned}
\log_{10}\lvert B(2k)\rvert &= \left(2k+\tfrac{1}{2}\right)\log_{10} (2k) - \frac{2k}{\log 10} + \tfrac{1}{2} \log_{10} (2\pi) + \log_{10} 2 - 2k\log_{10}(2\pi) + o(1)\\
&= \left(2k+\tfrac{1}{2}\right)\log_{10} (2k) - 2k\left(\log_{10} (2\pi) + \frac{1}{\log 10}\right) + \tfrac{1}{2} \log_{10}(8\pi) + o(1).
\end{aligned}$$
Now inserting $2k = 10^n$, that becomes
\begin{align}
\log_{10} \lvert B(10^n)\rvert &= \left(10^n + \tfrac{1}{2}\right)n - 10^n\left(\log_{10} (2\pi) + \frac{1}{\log 10}\right) + \tfrac{1}{2} \log (8\pi) + o(1)\\
&= 10^n\left(n - \log_{10}(2\pi) - \frac{1}{\log 10}\right) + \frac{n}{2} + \frac{1}{2} \log (8\pi) + o(1).
\end{align}
Since
$$\log_{10} (2\pi) + \frac{1}{\log 10} \approx 1.2324743502613666,$$
we expect indeed the first digits to form $n-2$, then followed by $76752564973...$ until the $\frac{n}{2}$ intervenes, which happens later and later for growing $n$.

(1) Which can be derived from the partial fraction decomposition of the cotangent, for example.
