Find Taylor series of: $x^x$ Find the Taylor seriespolynomial form of:

$$x^x$$
My attempt:

I started calculating derivatives of $x^x$ for the series,
$$f(x)=x^x$$
$$f'(x)=x^x(\ln x+1)$$
$$f''(x)=x^{x-1}+x^x(\ln x+1)$$
$$f'''(x)=\cdots$$
$$\vdots$$
But i could get only till certain terms and got frustrated,

Attempt no. 2:

$$f(x)=x^x$$
$$f(x)=e^{x\ln x}$$
$$f(x)=1+x\ln x+\frac{x^2\ln^2 x}{2!}+\cdots$$
$$f(x)=1+x({x-1}-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}+\cdots)+\frac{x^2}{2}({x-1}-\frac{(x-1)^2}{2}+\frac{(x-2)^3}{3}+\cdots)^2+\cdots$$
Now let's try to get $x^n$'s coefficient,
$$C(x^0)=1$$
In the bracket multiplied to x,
$$1+x(x-1-x^2+2x-1+x^3-3x+3x-1+\cdots)+\cdots$$
Following the lead
$$1+x(g(x)-1-1-1-1\cdots)+\cdots$$
$$C(x^1)=-1-1-1-\cdots$$
So,
$$C(x^1)=-\infty\cdots?$$
How do we write this equation?

Attempt 3:

$$f(x)=1+(x+1)\ln (x+1)+\frac{(x+1)^2}{2!}\ln^2(x+1)+\cdots$$
Now,
$$f(x)=1+(x+1)(x+\frac{x^2}{2}+\frac{x^3}{3}\cdots)+\cdots$$
Now,
$$C(x^0)=1$$
Considering $(1+x)^2,(1+x)^3\cdots$ contribution to$C(x^1)$
$$C(x^1)=1+1+1+1+\cdots$$
$$C(x^1)=\infty\cdots ?$$
Still.....stuck
 A: Hint: We can find an elaboration of the $n$-th derivative of $x^x$ in an example (p. 139) of Advanced Combinatorics by L. Comtet. The idea is based upon a clever Taylor series expansion. Using the  differential operator $D_x^j:=\frac{d^j}{dx^j}$ the following holds:

The $n$-th derivative of $x^x$ is
\begin{align*}
\color{blue}{D_x^n x^x=x^x\sum_{i=0}^n\binom{n}{i}(\ln(x))^i\sum_{j=0}^{n-i}b_{n-i,n-i-j}x^{-j}}\tag{1}
\end{align*}
  with $\color{blue}{b_{n,j}}$ the Lehmer-Comtet numbers.
These numbers follow the recurrence relation
  \begin{align*}
b_{n+1,j}=(j-n)b_{n,j}+b_{n,j-1}+nb_{n-1,j-1}\qquad\qquad n,j\geq 1
\end{align*}
  and the first values, together with initial values are listed below.
\begin{array}{c|cccccc}
n\setminus k&1&2&3&4&5&6\\
\hline
1&1\\
2&1&1\\
3&-1&3&1\\
4&2&-1&6&1\\
5&-6&0&5&10&1\\
6&24&4&-15&25&15&1\\
\end{array}
The values can be found in OEIS as A008296. They are called Lehmer-Comtet numbers and were stored in the archive by N.J.A.Sloane by referring precisely to the example we can see here.

Example: $n=2$
Let's look at a small example. Letting $n=2$ we obtain from (1) and the table with $b_{n,j}$:
\begin{align*}
D_x^2x^x&=x^x\sum_{i=0}^2\binom{2}{i}(\ln(x))^i\sum_{j=0}^{2-i}b_{2-i,2-i-j}x^{-j}\\
&=x^x\left(\binom{2}{0}\sum_{j=0}^2b_{2,2-j}x^{-j}+\binom{2}{1}\ln(x)\sum_{j=0}^1b_{1,1-j}x^{-j}\right.\\
&\qquad\qquad\left.+\binom{2}{2}\left(\ln(x)\right)^2\sum_{j=0}^0b_{0,0-j}x^{-j}\right)\\
&=x^x\left(\left(b_{2,2}+b_{2,1}\frac{1}{x}+b_{2,0}\frac{1}{x^2}\right)+2\ln(x)\left(b_{1,1}+b_{1,0}\frac{1}{x}\right)
+(\ln(x))^2b_{0,0}\right)\\
&=x^x\left(1+\frac{1}{x}+2\ln(x)+\left(\ln(x)\right)^2\right)
\end{align*} 
in accordance with the result of Wolfram Alpha.

Note: A detailed answer is provided in this MSE post.

