Q:

Taylor expansion of $x^x-1$ around 1

How do I find Taylor expansion of(around 1):
$$f(x)=x^x-1$$
The answer should be:
$$(x-1)+(x-1)^2+\frac 12(x-1)^3+\cdots$$
How the answer was obtained?

A:

Using the power series for $\log(1+x)$, we get
$$
\begin{align}
(1+x)\log(1+x)
&=(1+x)\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots\right)\\
&=x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{12}-\frac{x^5}{20}+\frac{x^6}{30}-\dots\tag{1}
\end{align}
$$
Plugging $(1)$ into the power series for $e^x$ yields
$$
(1+x)^{1+x}=1+x+x^2+\frac{x^3}{2}+\frac{x^4}{3}+\frac{x^5}{12}+\frac{3x^6}{40}-\frac{x^7}{120}+\dots\tag{2}
$$

Alternatively, use the binomial theorem to get
$$
(1+x)^{x+1}=1+\tfrac{x+1}{1}x+\tfrac{(x+1)x}{2}x^2+\tfrac{(x+1)x(x-1)}{6}x^3+\dots\tag{3}
$$
which also produces $(2)$.

Plug $x-1$ into $(2)$ and subract $1$ to get the Taylor Series for $x^x-1$ at $x=1$:
$$
x^x-1=(x-1)+(x-1)^2+\frac{(x-1)^3}{2}+\frac{(x-1)^4}{3}+\frac{(x-1)^5}{12}+\dots\tag{4}
$$

A:

Use the fact that the derivative of $x^x$ is $(\ln x + 1)x^x$, and expand around $x=1$:
$$y(x) = x^x,\ y'(x) =(\ln x + 1)x^x,\ y''(x)=\left((\ln x + 1)^2+\frac{1}{x}\right)x^x$$
$$y(1) = 1,\ y'(1) = 1,\ y''(x)=2$$
$$y(x) \sim 0 + (x-1)+(x-1)^2+...$$

