It is easy to prove:
$$e^{\ln\left(1+x\right)}=1+x$$
in algebraic way. However, how about prove it by Taylor expansion?
Know that: $$\ln\left(1+x\right)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\cdots\\e^z=1+z+\dfrac{z^2}{2}+\dfrac{z^3}{6}+\dfrac{z^4}{24}+\cdots$$ Then, $$e^{\ln\left(1+x\right)}=e^{x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots}=e^x\cdot e^{-\frac{x^2}{2}}\cdot e^\frac{x^3}{3}\cdot e^{-\frac{x^4}{4}}\cdots\\=\left(1+x+\dfrac{1}{2}x^2+\dfrac{1}{6}x^3+\cdots\right)\left[1-\dfrac{x^2}{2}+\dfrac{1}{2}\left(\dfrac{x^2}{2}\right)^2-\dfrac{1}{6}\left(\dfrac{x^2}{2}\right)^3+\cdots\right]\left[1+\dfrac{x^3}{6}+\dfrac{1}{2}\left(\dfrac{x^3}{6}\right)^2+\dfrac{1}{6}\left(\dfrac{x^3}{6}\right)^3+\cdots\right]\left[1+\dfrac{x^4}{24}+\dfrac{1}{2}\left(\dfrac{x^4}{24}\right)^2+\dfrac{1}{6}\left(\dfrac{x^4}{24}\right)^3+\cdots\right]$$ Till here, I am stuck. This expansion is very big that I can't really simplify. Though I find the coefficient of $x^0,x^1,x^2,x^3$ are $1,1,0,0$ respectively. However, I can't prove the higher degree coefficient. If there are someone can answer or giving tips, that's appreciated. Thank you!
Staying with $$e^{\ln\left(1+x\right)}=e^{x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}}$$ let $y=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}$ and use $$e^y=1+y+\frac 12 y^2+\frac 16 y^3+\frac 1{24} y^4$$ So, replacing $y$ and expanding $$y=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+O(x^5)$$ $$y^2=x^2-x^3+\frac{11 x^4}{12}-\frac{5 x^5}{6}+\frac{13 x^6}{36}-\frac{x^7}{6}+\frac{x^8}{16}$$ $$y^3=x^3-\frac{3 x^4}{2}+\frac{7 x^5}{4}-\frac{15 x^6}{8}+\frac{4 x^7}{3}-\frac{41 x^8}{48}+\frac{205 x^9}{432}-\frac{17 x^{10}}{96}+\frac{x^{11}}{16}-\frac{x^{12}}{64}$$ $$y^4=x^4-2 x^5+\frac{17 x^6}{6}-\frac{7 x^7}{2}+\frac{155 x^8}{48}-\frac{31 x^9}{12}+\frac{49 x^{10}}{27}-\frac{223 x^{11}}{216}+\frac{1355 x^{12}}{2592}-\frac{97 x^{13}}{432}+\frac{7 x^{14}}{96}-\frac{x^{15}}{48}+\frac{x^{16}}{256}$$ So, all of the above gives $$e^{x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}}=1+x-\frac{5 x^5}{24}-\frac{x^6}{72}-\frac{x^7}{144}+\cdots$$
Edit
What could be interesting is to define $$y_n=\sum_{k=1}^n (-1)^{k-1} \frac{x^k} k$$ to show that the expansion $$\exp\left(\sum_{k=0}^p \frac{y_n^k} {k!} \right)=1+x +a_{p+1} x^{p+1}+O(x^{p+2})$$ Starting at $p=2$, the sequence for the $a$ coefficients is $$\left\{-\frac{1}{2},\frac{5}{24},-\frac{5}{24},\frac{119}{720},-\frac{103}{720},\frac{5039}{40320},-\frac{40321}{362880},\frac{362879}{3628800},-\frac{329891}{3628 800},\frac{39916799}{479001600}\right\}$$