find
Clearly the top and bottom both tend to 0 as $\displaystyle x \to 1$.
So you can use L'Hospital's Rule.
$\displaystyle \lim_{x \to 1}\frac{x^n - 1}{x^p - 1} = \lim_{x \to 1}\frac{\frac{d}{dx}(x^n - 1)}{\frac{d}{dx}(x^p - 1)}$
$\displaystyle = \lim_{x \to 1}\frac{nx^{n - 1}}{px^{p - 1}}$
$\displaystyle = \frac{n}{p}$.
here's a hint:
we only need to show that $\displaystyle \forall a \in \mathbb{R} : \ \lim_{x\to1} \frac{x^a - 1}{x-1} = a.$ it's easy to prove it for rational numbers. now for any $\displaystyle n \in \mathbb{N}$ choose $\displaystyle q_n \in \mathbb{Q}$ such that $\displaystyle q_n - \frac{1}{n} \leq a \leq q_n + \frac{1}{n}.$ then $\displaystyle \lim_{n\to\infty} q_n = a$ and ...
Yeah yeah, and then $\displaystyle \lim_{x\to1}\frac{x^a-1}{x^p-1}=\lim_{x\to1}\left\{\frac{x^a-1}{x-1}\cdot\frac{x-1}{x^p-1}\right\}=\frac{a}{p}$
By the way, was your point that if we can prove that the above limit is true for any $\displaystyle a\in\mathbb{Q}$ and since any real number has a sequence of rational points which converges to it we are done?