The $n$'th derivative of $x^x$ I want to know the $n$'th derivative of $f(x)=x^x$.
Then, I'll calculate $f(0)$ with Taylor expansion of $f(x)$ on $a=1$.
Here is my answer, but it is unfinished.
The derivative of $f(x)=x^x$
$$\begin{align}
f'(x)&=x^x(\log x+1)\\
f''(x)&=x^x(\log x+1)^2+x^{x-1}\\
f'''(x)&=x^x(\log x+1)^3+3x^{x-1}(\log x+1)-x^{x-2}\\[5pt]
f(x)^{(4)}&=x^x(\log x+1)^4+4x^{x-1}(\log x+1)^2-4x^{x-2}(\log x+1)+3x^{x-2}+2x^{x-3}\\
f(x)^{(5)}&=x^x(\log x+1)^5+10x^{x-1}(\log x+1)^3-10x^{x-2}(\log x+1)^2+15x^{x-2}(\log x+1)\\&\quad+10x^{x-3}(\log x+1)-10x^{x-3}-6x^{x-4}\\
f(x)^{(6)}&=x^x(\log x+1)^6+15x^x(\log x+1)^4-20x^{x-2}(\log x+1)^3+45x^{x-2}(\log x+1)^2\\&\quad+30x^{x-3}(\log x+1)^2-50x^{x-3}(\log x+1)+15x^{x-3}-46x^{x-4}(\log x+1)\\&\quad+40x^{x-4}+24x^{x-5}
\end{align}$$
Taylor expansion of $f(x)=x^x$ in $a=1$
$$\begin{align}
f(x)&=\sum_{i=0}^{n-1}\frac{f^{(i)}(1)}{i!}\\[5pt]
&\qquad=\frac1{0!}+\frac1{1!}(x-1)+\frac2{2!}(x-1)^2+\frac3{3!}(x-1)^3+\frac8{4!}(x-1)^4+\frac{12}{5!}(x-1)^5\\&\qquad+\frac{54}{6!}(x-1)^6+\cdots
\end{align}$$

I tried the simplest approach to test myself. Let $u=x^{x-1}=e^{(x-1)\ln x}$, then $u=1+\sum_{n=1}^{\infty}\frac{1}{n!}(x-1)^n(\sum_{s=1}^{\infty}\frac{(-1)^{s-1}}{s}(x-1)^s)^n$ and expanding the binomial we have
$$u=1+\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{n!}C_{n,k}(x-1)^{n+k}$$
where
$$C_{n,k}=\sum_{r_1+r_2+...+r_n=k}^{r_i\geq 1,\forall i}\frac{(-1)^{k-n}}{r_1r_2...r_n}.$$
Letting $n+k=m$, we have $u=1+\sum_{m=2}^\infty a_m(x-1)^m$ where
$$a_m=\sum_{n=1}^{\lfloor m/2\rfloor}\frac{C_{n,m-n}}{n!}.$$
Then we have
$$y=x^x=1+(x-1)+(x-1)^2+\sum_{m=3}^{\infty}(a_m+a_{m-1})(x-1)^m.$$
$a_2=C_{1,1}=1$, $a_3=C_{1,2}=-\frac12$, $a_4=C_{1,3}+\frac{C_{2,2}}{2}=\frac 13+\frac12=\frac56$, $a_5=C_{1,4}+\frac{C_{2,3}}{2}=-\frac 14+-\frac12=-\frac34$, $a_6=C_{1,5}+\frac{C_{2,4}}{2}+\frac{C_{3,3}}{6}=\frac15+\frac{11}{24}+\frac16=\frac{33}{40},.....$ and hence
$$y=1+(x-1)+(x-1)^2+\frac{1}{2}(x-1)^3+\frac{1}{3}(x-1)^4+\frac{1}{12}(x-1)^5+\frac{3}{40}(x-1)^6+...$$
