An integral involving Sophomore's dream constant

The Sophomore's dream constant denoted as $\mathscr{S}$ is defined to be the series:

$$\mathscr{S}= \sum_{n=1}^{\infty} \frac{1}{n^n}$$

One interesting identity is the following

$$\int_0^1 x^{-x}\, {\rm d}x = \sum_{n=1}^{\infty} n^{-n} = \mathscr{S} \tag{1}$$

In this thread however we take the above for granted (and leave the proof to the reader since it is not that difficult)  and prove another interesting identity which is:

$$\int_0^1 \frac{\ln x}{x^x}\, {\rm d}x =- \mathscr{S}$$

Solution
 We have successively:

$$\begin{aligned}
\int_{0}^{1} \frac{\ln x}{x^x}\, {\rm d}x &=\int_{0}^{1}x^{-x} \ln x \, {\rm d}x \\
 &= \int_{0}^{1}e^{-x \ln x} \ln x \, {\rm d}x\\
 &= \int_{0}^{1}\ln x \sum_{n=0}^{\infty} \frac{\left ( -x \right )^n \left ( \ln x \right )^n}{n!}\, {\rm d}x \\
 &= \sum_{n=0}^{\infty}\int_{0}^{1}\frac{(-x)^n \left ( \ln x \right )^{n+1}}{n!}\, {\rm d}x\\
 &\overset{\ln x = - \frac{t}{n+1}}{=\! =\! =\! =\! =\! =\! =\!} \sum_{n=0}^{\infty}\int_{0}^{\infty}\frac{(-1)^{2n+1} e^{-t}t^{n+1}}{\left ( n+1 \right )^{n+2} n!} \\
 &= -\sum_{n=0}^{\infty} \frac{\Gamma (n+2)}{\left ( n+1 \right )^{n+2}n!}\\
 &= - \sum_{n=0}^{\infty} \frac{(n+1)!}{\left ( n+1 \right )^{n+2}n!} \\
 &= - \sum_{n=1}^{\infty} \frac{1}{n^n} \\
 &= -\mathscr{S}
\end{aligned}$$

and the calculations are complete.

$(*)$ $\Gamma(n+2)=(n+1)!$ is taken for granted since the $\Gamma$ function is a well studied function. Informations on the function can found everywhere on the Net.