We are given to find:
$$L=\lim_{x\to1}\frac{x^m-1}{x^n-1}$$ where $$m,n\in\mathbb{N}$$
Using the hint I suggested, we may write:
$$x^m-1=(x-1)\sum_{k=0}^{m-1}\left(x^k\right)$$
$$x^n-1=(x-1)\sum_{k=0}^{n-1}\left(x^k\right)$$
And so our limit becomes:
$$L=\lim_{x\to1}\frac{(x-1)\sum\limits_{k=0}^{m-1}\left(x^k\right)}{(x-1)\sum\limits_{k=0}^{n-1}\left(x^k\right)}=\lim_{x\to1}\frac{\sum\limits_{k=0}^{m-1}\left(x^k\right)}{\sum\limits_{k=0}^{n-1}\left(x^k\right)}=\frac{\sum\limits_{k=0}^{m-1}\left(1\right)}{\sum\limits_{k=0}^{n-1}\left(1\right)}=\frac{\sum\limits_{k=1}^{m}\left(1\right)}{\sum\limits_{k=1}^{n}\left(1\right)}=\frac{m}{n}$$