How to show that $x=e^y$, where $y=\sum \limits_{n=1}^\infty \frac{(-1)^{n-1}}{n} (x-1)^n$, not using logarithm? Let $|x-1|<1$ and $y= \displaystyle \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} (x-1)^n$. 
How to show, without using the fact that $y=\ln x$, but using properties of absolutely convergent series, that $e^y=x$?
 A: Given a power series, you can integrate (or) differentiate term by term within its radius of convergence, where it converges absolutely. The series
$$
\begin{align}
y = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} (x-1)^n
\end{align}
$$
converges absolutely for $|x-1| < 1$. Hence, term by term differentiation gives us 
$$
\begin{align}
\frac{dy}{dx} = \sum_{n=1}^{\infty} (-1)^{n-1} (x-1)^{n-1} = \sum_{n=1}^{\infty} (1-x)^{n-1} = \frac1{1-(1-x)} = \frac1x
\end{align}
$$
for $|x-1| < 1$. Now integrate to get that $y = \log(x) + c$. Set $x=1$ in the initial given equation to get that $y=0$ and hence $c=0$. Hence, we get that, $$x = e^y$$
A: Define
$$
f(x)=-\sum_{n=1}^\infty\frac{(1-x)^n}{n}\tag{1a}
$$
so that
$$
f(1-x)=-\sum_{n=1}^\infty\frac{x^n}{n}\tag{1b}
$$
After which, your question becomes: show that $x=e^{f(x)}$.
$$
\begin{align}
f((1-u)(1-v))
&=f(1-(u+v-uv))\\
&=-\sum_{n=1}^{\infty}\frac{(u+v-uv)^n}{n}\\
&=-\sum_{n=1}^{\infty}\frac{(u(1-v)+v)^n}{n}\\
&=-\sum_{n=1}^{\infty}\sum_{k=0}^n\frac{1}{n}\binom{n}{k}(u(1-v))^kv^{n-k}\\
&=-\sum_{n=1}^\infty\frac{1}{n}v^n-\sum_{k=1}^\infty\sum_{n=k}^\infty\frac{1}{k}\binom{n-1}{k-1}(u(1-v))^kv^{n-k}\\
&=-\sum_{n=1}^\infty\frac{1}{n}v^n-\sum_{k=1}^\infty\frac{1}{k}\left(\frac{u(1-v)}{1-v}\right)^k\\
&=-\sum_{n=1}^\infty\frac{1}{n}v^n-\sum_{k=1}^\infty\frac{1}{k}u^k\\
&=f(1-u)+f(1-v)\tag{2}
\end{align}
$$
Thus, $(2)$ says that
$$
f(xy)=f(x)+f(y)\tag{3}
$$
Equation $(3)$ insures that
$$
f(1)=f(x\cdot1)-f(x)=0\tag{4}
$$
Equations $(3)$ and $(4)$ yield
$$
\begin{align}
f'(x)
&=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\
&=\lim_{h\to0}\frac{f(x(1+h/x))-f(x)}{h}\\
&=\frac{1}{x}\lim_{h\to0}\frac{f(1+h/x)-f(1)}{h/x}\\
&=\frac{1}{x}f'(1)\tag{5}
\end{align}
$$
Let $g(x)=e^{f(x)}$, then $g(1)=1$ and
$$
\begin{align}
g'(x)
&=g(x)f'(x)\\
&=g(x)\frac{1}{x}f'(1)\tag{6}
\end{align}
$$
Note that equation $(6)$ implies
$$
\begin{align}
x^{f'(1)+1}\left(x^{-f'(1)}g(x)\right)'
&=-f'(1)g(x)+xg'(x)\\
&=0\tag{7}
\end{align}
$$
and equation $(7)$ implies
$$
x^{-f'(1)}g(x)=C\tag{8}
$$
Since $g(1)=1$, $C=1$, and therefore,
$$
g(x)=x^{f'(1)}\tag{9}
$$
Referring back to equation $(1)$, we see that $f'(1)=1$, so we have
$$
g(x)=x\tag{10}
$$
as requested.
