Determining the value of $\sum_{n=2}^\infty \frac{1}{n^n-1}$ $$\displaystyle\sum_{n=1}^\infty\sum_{m=2}^\infty \frac{1}{m^{mn}}=\sum_{n=1}^\infty\left(\frac{1}{2^{2n}}+\frac{1}{3^{3n}}+\frac{1}{4^{4n}}+\cdots\right) \tag{$\star$}$$
I have a strong suspicion that the above summation converges, although I'm not sure how to prove it. But I'm more interested in the precise value of $(\star)$, especially a nice representation (if possible).
Here's what I have so far:
\begin{align}
\sum_{n=1}^\infty\left(\frac{1}{2^{2n}}+\frac{1}{3^{3n}}+\frac{1}{4^{4n}}+\cdots\right) & = \sum_{n=1}^\infty\frac{1}{2^{2n}}+\sum_{n=1}^\infty\frac{1}{3^{3n}}+\sum_{n=1}^\infty\frac{1}{4^{4n}}+\cdots \\ & = \sum_{n=1}^\infty\frac{1}{\left(2^{2}\right)^n}+\sum_{n=1}^\infty\frac{1}{\left(3^{3}\right)^n}+\sum_{n=1}^\infty\frac{1}{\left(4^{4}\right)^n}+\cdots \\ & = \frac{1}{2^2-1}+\frac{1}{3^3-1}+\frac{1}{4^4-1}+\cdots\\ & =\sum_{n=2}^\infty \frac{1}{n^n-1} 
\end{align}
So $(\star)=\displaystyle\sum_{n=2}^\infty \frac{1}{n^n-1}$. From here, all I know how to do is get an approximate value of the solution. How would you go about finding its exact value?


 A: If you want more figures $$\displaystyle \sum_{n=2}^\infty \frac{1}{n^n-1}=0.37605925334160927467565605197313357724916639675$$ All the digits are obtained summing up to $31$.
An amazing approximation of it is
$$\frac{76 \pi ^2+651 \pi-563 }{308 \pi ^2+481 \pi+1385}$$ which is in a relative error of $2.56\times 10^{-18}\text{ %}$.
Edit
Using exact arithmetic, I computed the value for $1000$ exact decimal places (if you want the number, tell me). The result is obtained summing up to $n=386$.
Whet looked interesting (at least to me) is that, negelcting the $-1$ in denominator
$$a_n=\frac{1}{n^n}\implies \frac{a_{n+1}}{a_n}=\frac{1}{e n}-\frac{1}{2 e n^2}+O\left(\frac{1}{n^3}\right)$$ So, writing
$$\sum_{n=2}^\infty \frac{1}{n^n-1}=\sum_{n=2}^{k-1} \frac{1}{n^n-1}+\sum_{n=k}^\infty \frac{1}{n^n-1}=\sum_{n=2}^{k-1} \frac{1}{n^n-1}+e^{\frac{1}{2 e}}I_0\left(\frac{1}{2 e}\right)\frac{ k }{k^k-1}$$
So, if we want the remainder to be less than $\epsilon$ we need
$$k\sim\frac{\log \left(\frac{b}{\epsilon }\right)}{W\left(\log \left(\frac{b}{\epsilon}\right)\right)}\qquad \text{where} \qquad b=e^{\frac{1}{2 e}}I_0\left(\frac{1}{2 e}\right)$$ Applied to the case where $\epsilon =10^{-1000}$, this gives $k=386.55$.
