<s>[INST] Find the power series expansion of $f(x)= \frac{e^x - 1}{x}$ 
Prove that the following function is analytical in 0 and find its power series centered in 0
$$f(x)= \frac{e^x - 1}{x}, f(0)=1$$

I'm trying to write $f$ as some kind of combination of function with known Taylor series expansion in an open interval around zero, which would prove the analicity:
$$f(x)= \frac{e^x}{x}- \frac{1}{x}=\frac{1}{x}\sum_{n=0}^{\infty} \frac{x^n}{n! } - \frac{1}{x} =\sum_{n=0}^{\infty} \frac{x^{n-1}}{n! } - \frac{1}{x}$$
but I don't know about any power series expansion of $\frac{1}{x}$
 [/INST] You know that $e^x= \sum_{n=0}^\infty \frac{x^n}{n!}= 1+ x+ \frac{x^1}{2!}+ \cdot\cdot\cdot+ \frac{x^n}{n!}+ \cdot\cdot\cdot$,  so 
$$
e^x- 1= \sum_{n=1}^\infty \frac{x^n}{n!}= x+ \frac{x^1}{2!}+ \cdot\cdot\cdot+ \frac{x^n}{n!}+ \cdot\cdot\cdot.
$$ 
thus 
$$
\frac{e^{x}- 1}{x}= \sum_{n= 1}^\infty \frac{x^{n-1}}{n!}=\sum_{n= 0}^\infty \frac{x^n}{(n+1)!}= 1+ \frac{x}{2!}+ \frac{x^2}{3!}+ \cdot\cdot\cdot+ \frac{x^n}{n!}+ \cdot\cdot\cdot
$$
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