Expanding $x^x$ to series with $o(x)$ polynomials I have some doubts in finding $x^x$ series.  I know that from Taylor theorem I have
$$ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(k)}(a)}{k!}(x-a)^k + o(x^k) $$
I want to have $o(x)$ polynomials so:
Let $$ f(x) = x^x $$
$$ f'(x) = (e^{x\cdot \ln x}) = (\ln x + 1)x^x$$
In use of taylor I should
$$ f(x) = f(0) + \frac{f'(0)(x)}{1!} + o(x) $$
but $f(0)$ is not defined $0^0$. Also $f'(0)$ is not defined...
But I found informations that it should be 
$$  1+ \ln x \cdot x + o(x)$$
I don't know why :-(

As you have written
$$x^x=e^{x\ln{x}}$$
and as the series expansion of $e^x$ converges for all values of $x\in\mathbb{C}$ one can write
$$e^x=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+...$$
$$e^{x\ln{x}}=\frac{(x\ln{x})^0}{0!}+\frac{(x\ln{x})^1}{1!}+\frac{(x\ln{x})^2}{2!}+...$$
$$\therefore x^x=1+x\ln{x}+\frac{x^2\ln^2{x}}{2}+\frac{x^3\ln^3{x}}{6}+...$$
