Show that $\int\limits_0^1 \left(x^{x}\right)^{\left(x^{x}\right)^{\left(x^{x}\right)^{\left(x^{x}\right)^{⋰}}}}\ \mathrm{d}x=\frac{\pi^2}{12}$.

How can it be shown that $$\lim_{p\to\infty}I(p)= \lim_{p \to \infty}\int^{1}_0 (x^x)^{\scriptscriptstyle {(x^x)^{(x^x)^{(x^x)^{(x^x)^{(x^x)...(p \; times)}}}}}} dx= \frac{\pi^2}{12}$$

$I(1)=\int^{1}_0 x^x dx=\sum_{n=0} ^{\infty} \frac{1}{n!}\int^{1}_0 \ln(x)^nx^n dx$

also $\int^{1}_0 \ln(x)^nx^n dx=(-1)^n (1+n)^{-1-n}n!$ .

I don't know how to calculate $I(p)$ beyond $p=1.$

Note: This integral was proposed on Romanian Mathematical Magazine and two solutions can be found here.

From here: $$W(x)=\sum _{n=1}^{\infty } \frac{(-n)^{n-1} x^n}{n!}$$

and with CAS help: $$\begin{align} \int_0^1 -\frac{W(-x \ln (x))}{x \ln (x)} \, dx &=\int_0^1 \left(\sum _{n=1}^{\infty } \frac{(-1)^{2 n} n^{-1+n} x^{-1+n} \ln ^{-1+n}(x)}{n!}\right) \, dx \\ &=\sum _{n=1}^{\infty } \int_0^1 \frac{(-1)^{2 n} n^{-1+n} x^{-1+n} \ln ^{-1+n}(x)}{n!} \, dx\\ &=\sum _{n=1}^{\infty } \frac{\left((-1)^{2 n} n^{-1+n}\right) \int_0^1 x^{-1+n} \ln ^{-1+n}(x) \, dx}{n!}\\ &=\sum _{n=1}^{\infty } \frac{\left((-1)^{2 n} n^{-1+n}\right) \left(-\left(-\frac{1}{n}\right)^n \Gamma (n)\right)}{n!}\\ &=\sum _{n=1}^{\infty } \frac{(-1)^{1+3 n}}{n^2}\\ &=\frac{\pi ^2}{12}. \end{align}$$