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

Extract from my note Derivatives of generalized power functions that appeared in the Reader Reflections column of Mathematics Teacher [Volume 103, Number 9; May 2010; pp. 630-631]:

Regarding the editor's note on higher derivatives of $x^{x},$ let $n$ be a positive integer, ${n \choose k}$ be the usual binomial coefficient, $f(x) = x^{x},$ and $g(x) = 1 + \ln{x}.$ Kulkarni (1984) gave the recursion formula
$$ f^{(n+1)}(x) \;\; = \;\; f^{(n)}(x)g(x) \;\; + \;\; \sum\limits_{k=1}^{n} \left[ {n \choose k} (-1)^{k-1}(k-1)! \right] f^{(n-k)}(x)x^{-k} $$
by observing $f^{(n+1)}(x)$ is the $n$th derivative of $f'(x) = x^x(1 + \ln{x})$ and then writing down the Leibniz formula for the $n$th derivative of a product, using the fact that the $k$th derivative of $1 + \ln x$ is $(-1)^{k-1}(k-1)!x^{-k}.$
S. B. Kulkarni, Solution to Problem 3977, School Science and Mathematics 84 #7 (November 1984), 629-630.

