Taylor series of the function $f(x) = (1+x) ^{\frac{1}{x}}$ Good night!! I got this problem and I'd like to find all the mistakes in my statement. This is my prblem.
Find the binomial coefficients of $f(x) = (1+x)^{\alpha}$, with $\alpha \in \mathbb{R}$, and then find the Taylor serie of $f(x) = (1+x)^{\frac{1}{x}}$ around $x = 0$
What I try to do: I rewrite $f$ as $f(x) = e^{\alpha\ln\,(1+x)}$, and derivating I found that
$$ f^{(n)}(x) = \alpha(\alpha - 1)(\alpha - 2)\dots(\alpha - (n-1))e^{\alpha\ln\,(1+x)}\left( \dfrac{1}{1+x} \right)^n $$
Question: Could I derive like this: $f'(x) = \alpha(1+x)^{\alpha - 1}$?
Later, I used this result to show that
$$ (1+x)^{\frac{1}{x}} = e + \alpha(\alpha - 1)(\alpha - 2)\cdots(\alpha - (n-1)) $$
I add $e$ because
$$ \lim_{x \rightarrow 0}\,\dfrac{e - (1+x)^{\frac{1}{x}}}{x} = e $$
What is wrong with my argument? Thanks for your help!!!

Hint
For building, around $x=0$, Taylor expansion of
$$f= (1+x) ^{\frac{1}{x}}$$ I suggest you first write $$\log(f) =\frac{1}{x}\log (1+x)$$ Use the development of $\log(1+x)$ and get $$\log(f) =1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\frac{x^4}{5}-\frac{x^5}{6}+O\left(x^6\right)$$ Now, define $$y=-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\frac{x^4}{5}-\frac{x^5}{6}+...$$ So $$f=e^{\log(f)}=e~~~e^y$$ Use Taylor series of $e^y$ around $y=0$ and in the result, replace $y$ by its value and expand. You should arrive to $$(1+x) ^{\frac{1}{x}} \simeq e\Big(1-\frac{x}{2}+\frac{11  x^2}{24}-\frac{7 x^3}{16}+\frac{2447 
   x^4}{5760}-\frac{959  x^5}{2304}+\frac{238043  x^6}{580608}+..\Big)$$ where the coefficients are those of $OEIS$ sequences $A055505$ and $A055535$ as 
Robert Israel mentioned.
Added later to this answer
Another way could be to write $$f =e^{\frac{1}{x}\log (1+x)}=e^{1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\frac{x^4}{5}-\frac{x^5}{6}+O\left(x^6\right)}=e~e^{-\frac{x}{2}}~e^{\frac{x^2}{3}}~e^{-\frac{x^3}{4}}~e^{\frac{x^4}{5}}...$$ to use for each term the first terms of the series of $e^z$ and to perform the multiplications. It is probably simpler than in my initial answer.
