If $\mathfrak{B}$ is a banach algebra and $f\in\mathfrak{B}$ such that $\lVert 1-f\rVert<1$ then $f$ is in $\exp\mathfrak{B}$

Question

I'm reading the book Banach Algebra Techniques in operator Theory by Ronald G. Douglas.

There it is defined that $g=\sum\limits_{n=1}^\infty \frac{-1}{n}(1-f)^n$. It is convergent absolutely as $\lVert 1-f\rVert<1$. Then we are asked to substitute the value of $g$ in the expansion of $\exp g=\sum\limits_{n=0}^\infty \frac{g^n}{n!}$. But directly substituting $g$ making the computations more complicated, specifically how to compute $g^n$? But I have observed the following-

$\exp g=1+g+\sum\limits_{n\ge2} \frac{g^n}{n!}=1+(-1+f)-\sum\limits_{n\ge2}\frac{(1-f)^n}{n}+\sum\limits_{n\ge2}\frac{g^n}{n!}$. So $\exp g=f$ if and only if $\sum\limits_{n\ge2}\frac{g^n}{n!}=\sum\limits_{n\ge2}\frac{(1-f)^n}{n}$ if and only if $\sum\limits_{n\ge 2}\frac{g^n-(n-1)!(1-f)^n}{n!}=0$. But I cannot prove this last expression.

Can anyone provide me a wayout to finish the proof? Thanks for your help in advance.