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 $a=1$
$$\begin{align}
f(x)&=\sum_{i=0}^{n-1}\frac{f^{(i)}(1)}{i!}\\[5pt]
&\qquad=\frac1{0!}+\frac1{1!}(x-1)+\frac2{2!}(x-1)^2+\frac3{3!}(x-1)^3+\frac8{4!}(x-1)^4+\frac{12}{5!}(x-1)^5\\&\qquad+\frac{54}{6!}(x-1)^6+\cdots
\end{align}$$

Hint
I think that you could simplify the calculations using sucessively logarithmic derivatives. This means that you will cascade the expressions and the $n^{th}$ derivative of $f(x)$ will be expressed as function of lower order derivatives. 
In fact I suppose that some of your derivatives are not correct since, after simplications, we get
$$f(x)=1+(x-1)+(x-1)^2+\frac{1}{2} (x-1)^3+\frac{1}{3} (x-1)^4+\frac{1}{12}
   (x-1)^5+\frac{3}{40} (x-1)^6-\frac{1}{120} (x-1)^7+\frac{59
   (x-1)^8}{2520}-\frac{71 (x-1)^9}{5040}+\frac{131
   (x-1)^{10}}{10080}+O\left((x-1)^{11}\right)$$
In order to check you expansion, I suggest you put is on the same graph as the original function plotting for $0 \leq x \leq 2$; the two curves are supposed to be very similar.
