Show that it is equal to the integral

Question

How can we show that $$\sum_{n=1}^{+\infty} \frac{1}{n^n} = \int_0^1 \frac{1}{x^x}dx$$

Could you give me some hints??

Answer 1

Hint: Write $x^{x} = e^{x\ln x}$, use $$e^{x} = \sum_{k=0}^{\infty}\frac{x^k}{k!}$$

and integrate term by term.