The rth term in $(1+x)^{1/x} = e[1 - \frac {1}{2}x + \frac {11}{24}x^2 - \frac {7}{16}x^3 + ....]$ Let 
$$(1+x)^{\frac {1}{x}} = e.G(x)$$
Taking logarithm on both sides,
$$\frac {1}{x} \log {(1+x)} = 1 + \log {G(x)}$$
Putting in the Taylor expansion for $\log {(1+x)}$ we have,
$$\frac {1}{x}(x - \frac {1}{2}x^2 + \frac {1}{3}x^3 - ....) = 1 + \log {G(x)}$$
Solving for $G(x)$ we have,
$$G(x) = e^{-\frac {1}{2}x + \frac {1}{3}x^2 - ...}$$
The difficulty starts here, for in order to get the desired Taylor expansion $e[1 - \frac {1}{2}x + \frac {11}{24}x^2 - \frac {7}{16}x^3 + ....]$ I have to plug in the entire expansion $-\frac {1}{2}x + \frac {1}{3}x^2 - ...$ for the variable in the Taylor expansion for $e$. Is there any alternative way? My main question is : what is the rth term of the desired expansion $e[1 - \frac {1}{2}x + \frac {11}{24}x^2 - \frac {7}{16}x^3 + ....]$ in its closed form?

\begin{align}
\dfrac1e(1+x)^{1/x} 
&= \exp\left(\dfrac{\ln(1+x)-x}{x}\right) \\
&= \exp\left(-\dfrac{x}{2}+\dfrac{x^2}{3}-\dfrac{x^3}{4}+\cdots\right) \\
&= 1+\left(-\dfrac{x}{2}+\dfrac{x^2}{3}-\dfrac{x^3}{4}+\cdots\right)+\dfrac12\left(-\dfrac{x}{2}+\dfrac{x^2}{3}-\dfrac{x^3}{4}+\cdots\right)^2 \\
& +\dfrac16\left(-\dfrac{x}{2}+\dfrac{x^2}{3}-\dfrac{x^3}{4}+\cdots\right)^3+\cdots\\
&= 1-\frac{1}{2}x+\left(\dfrac13+\dfrac18\right)x^2+\left(-\dfrac14+\dfrac122\dfrac{-1}{2}\dfrac13+\dfrac16\dfrac{-1}{8}\right)x^3+\cdots
\end{align}
