So you are trying to prove that your function is equal to its power series representation, and as you said to do this you will use Taylors theorem and inequality.
$\mathbf{Thereom:}$ If $f(x)=T_n(x)+R_n(x) , $ where $T_n$ is the $n^{th}$ degree taylor polynomial of $f$ at $a$ and $\lim_{n\to \infty} R_n(x)=0$ for $|x-a| \lt R$ ,
then $f$ is equal to the sum of its Taylor series on the interval $|x-a| \lt R$ ( theorem as given in Stewart 7ed.)
And we are wanting to take advantage of the following
$\mathbf{Lemma}:$ If $| f^{n+1}(x)| \le M$ for $|x-a| \le d$ , then the remainder $R_n(x)$ of the Taylor series satisfies the inequality
$$|R_n(x)| \le \frac{M}{(n+1)!}|x-a|^{n+1}$$ for |$x-a| \le d$
So what is the general form of the taylor series for $f(x)=\frac{1}{x}$ centred at $x=1$?
As you said , it has the general form $\sum_{n=0}^{\infty}(-1)^{n}(x-1)^{n}$ and converges for $|1-x| \lt 1$
Now , $f(x)=(1/x)$ and we have that $f^{n}(1)=(-1)^{n}n!$
$f^{n+1}(1)=(-1)^{n+1}(n+1)! \le (n+1)! $
$|x-a| \le 1$
and by the lemma we have $| R_n(x)| \le \frac{(n+1)!}{(n+1)!}|x-a|^{n+1}$
Do you see anything we could do from here?
In order to get the power series expansion we could multiply the series $e^z$ with $\frac{1}{1+z}$ using the Cauchy product
We obtain
\begin{align*}
\frac{e^z}{1+z}&=\left(\sum_{k=0}^{\infty}\frac{z^k}{k!}\right)\left(\sum_{l=0}^\infty (-z)^l\right)\tag{1}\\
&=\sum_{n=0}^\infty\left(\sum_{{k+l=n}\atop{k,l\geq 0}}\frac{(-1)^l}{k!}\right)z^n\tag{2}\\
&=\sum_{n=0}^{\infty}\left(\sum_{k=0}^n\frac{(-1)^{n-k}}{k!}\right)z^n
\end{align*}
Comment:
We observe the function
\begin{align*}
\frac{1}{1+z}
\end{align*}
has a simple pole at $z=-1$. We also know that the exponential function is an entire function, i.e. analytic in $\mathbb{C}$. Since the radius of convergence is the distance from the center $z=0$ to the nearest singularity, we conclude the radius is $1$.
Best Answer
$$e^x- 1 = \sum_{n=1}^\infty \frac{x^n}{n!}$$ Now divide each term by $x$.