prove that $\displaystyle\int_0^1 \dfrac{1}{x^x}dx = 1 + \dfrac{1}{2^2} + \dfrac{1}{3^3} + \dfrac{1}{4^4} + \ldots $ $\displaystyle\int_0^1 \dfrac{1}{x^x}dx = 1 + \dfrac{1}{2^2} + \dfrac{1}{3^3} + \dfrac{1}{4^4} + \ldots $
the only idea I have is using the series expansion of $x^{-x} \approx 1 - x\log x + \dfrac{(x\log x)^2}{2!} - \ldots$. But it ends up little complicated, any idea?

Yes go ahead
$$S=\int_{0}^{1} x^{-x}  dx= \int_{0}^{1} e^{-x\ln x} dx =\int_{0}^{1} \sum_{k=0}^{\infty} \frac{(-x \ln x)^k}{k!}dx.$$
Let $x=e^{-t}$,and use $\int_{0}^{\infty} t^n e^{-at}dt=\frac{n!}{a^{n+1}}.$
$$\implies S=\sum_{k=0}^{\infty} \int_{0}^{\infty}\frac{t^ke^{-(k+1)t}}{k!} dt=\sum_{k=0}^{\infty}\frac{1}{(k+1)^{k+1}}$$
