For this exercise we can use that $e^x = \sum\limits_{n=0}^{\infty}\frac{x^n}{n!}$ converges uniformly in all limited subset of $\mathbb{R}$ and have to show that $x^{x} = \sum\limits_{n=0}^{\infty}\frac{x^n \ln(x)^n}{n!}$ converges uniformly for all $x \in [0,1]$. I can show that it converges uniformly for all $x \in (0,1)$ by Dirichlet Test and $x=1$ by particular case, but I don't know if it is really true for $x = 0$. Is it possible to converge uniformly even having $ln(0)$ not existing? (As for $0^0$, we have a note saying to consider that as $1$).
Also, the next question is to show that $\int _0^1 \! x^x \, \mathrm{d}x = \sum\limits_{n=0}^{\infty} \frac{1}{n!}\int _0^1\! {x^n\ln(x)^n} \, \mathrm{d}x$. This would be easy if I had uniform convergence for $x = 0$, because I could use a Theorem that says that I can interchange sum and integral when convergence is uniform for all $x$ in integral limits.