Finding a limit without using L'Hospital's rule I encountered an issue for finding the limit for a function, one small step I just can't get. 
I know that this is classic case for L'Hospital's however our calculus course didn't reach it and I can not use it for this one.
$
2.3 .m,n\in\mathbb{N}\quad\underset{x\rightarrow1}{\lim}\frac{x^{m}-1}{x^{n}-1}$
$x\neq1$
\begin{aligned}\underset{x\rightarrow1}{\lim}\frac{x^{m}-1}{x^{n}-1}=\underset{x\rightarrow1}{\lim}\frac{x^{m}-1^{m}}{x^{n}-1^{n}}=\underset{x\rightarrow1}{\lim}\frac{\left(x-1\right)\sum_{i=0}^{m-1}x^{i}\cdot1^{m-i-1}}{\left(x-1\right)\sum_{i=0}^{m-1}x^{j}\cdot1^{n-j-1}}=\\
\underset{x\rightarrow1}{\lim}\frac{\left(m-1\right)x^{m-1}}{\left(n-1\right)x^{n-1}}=\underset{x\rightarrow1}{\lim}\frac{mx^{m-1}-x^{m-1}}{nx^{n-1}-x^{n-1}}
\end{aligned}
And here I stuck... 
know I need to reach $\frac{m}{n}$ but how can I proceed...?!!? nothing to cancel out here... 

$\lim_{x\to 1} \dfrac{x^m - 1}{x^n - 1} =
\lim_{x\to 1} \dfrac{(x-1)(x^{m - 1} + x^{m - 2} + x^{m - 3} + ... + 1)}{(x-1)(x^{n - 1} + x^{n - 2} + x^{n - 3} + ... + 1)} =
\lim_{x\to 1} \dfrac{(x-1)\sum_{i=1}^{m} x^{m-i}}{(x-1)\sum_{j=1}^{n} x^{n-j}} =
\lim_{x\to 1} \dfrac{\sum_{i=1}^{m} x^{m-i}}{\sum_{j=1}^{n} x^{n-j}} =
\dfrac{\sum_{i=1}^{m} 1^{m-i}}{\sum_{j=1}^{n} 1^{n-j}} =
\dfrac{\sum_{i=1}^{m} 1}{\sum_{j=1}^{n} 1} = 
\dfrac{m}{n}$
