I think I've found the antiderivative of $x^x$

Question

If you do integration by parts for $x^x$, while repeatedly using $dx$ as $dv$ and $x^x$ for $u$, you get an expansion that looks like $$\int x^xdx=xx^x -\int xx^x(1+\ln x)dx$$ Because we can find $x^x$ as many times as we want, we can keep solving the integral with dv as some power of x, we can expand to something like: $$\int x^xdx = xx^x-\frac{x^2}{2} \frac{d}{dx}x^x+\frac{x^3}{6}\frac{d}{dx}\frac{d}{dx}x^x-...=\sum^\infty_{n=1}{(-1)^{n+1} \left(\frac{x^n}{n!}\right)\left(\frac{d^{n-1}}{(dx)^{n-1}}\right)}$$ This seems right on paper but I think I've made a mistake somewhere because I've been told everywhere that there is no antiderivative for $x^x$