<s>[INST] Find $f^{(n)}(0) \text{ for } n = 1,2,3,...$ Find $f^{(n)}(0) \text{ for } n = 1,2,3,...$ where 
$$ f(x) = \begin{cases}
\frac{e^x - 1}{x}, & \text{when } x \neq 0 \\
1, & \text{when } x = 0 \\
\end{cases} $$
My approach
I decided to calculate some first $f$ derivatives. In that case I defined:
$$ g(x) = \frac{e^x - 1}{x} \text{ for } x\neq0$$
$$g'(x) = ...= \frac{e^x x-e^x+1}{x^2} $$
$$g''(x) = \text{..a lot of calculus..} =\frac{e^x x^2-2 e^x x+2 e^x-2}{x^3} $$
$$ g^{(3)}(x) = ... = \frac{e^x x^3-3 e^x x^2+6 e^x x-6 e^x+6}{x^4}  $$
$$ g^{(4)}(x) = \frac{e^x x^4-4 e^x x^3+12 e^x x^2-24 e^x x+24 e^x-24}{x^5}$$
But I don't see a pattern. Some of last factors are $n!$. Denominator is $x^{n-1}$. But for the rest I haven't got idea.

I know also that for $n>0$
$$ f^{(n)}(x) = \begin{cases}
\frac{e^x - 1}{x}, & \text{when } x \neq 0 \\
0, & \text{when } x = 0 \\
\end{cases} $$
so theoretically the answer is just $ 0 $ but I am not sure if solution can be so simple...
 [/INST] Since$$f(x)=1+\frac x{2!}+\frac{x^2}{3!}+\cdots+\frac{x^n}{(n+1)!}+\cdots,$$then$$(\forall n\in\mathbb{Z}^+):\frac{f^{(n)}(0)}{n!}=\frac1{(n+1)!}$$and therefore$$(\forall n\in\mathbb{Z}^+):f^{(n)}(0)=\frac1{n+1}.$$
</s>