Evaluate $\lim_{x\to 1}\frac{(x^n -1)(x^{n-1}-1)\cdots(x^{n-k+1}-1)}{(x-1)(x^2-1)\cdots(x^k-1)}$ 
Evaluate:$$
\lim_{x\to 1}\frac{(x^n -1)(x^{n-1}-1)\cdots(x^{n-k+1}-1)}{(x-1)(x^2-1)\cdots(x^k-1)}
$$

I'm trying to spot an error in my calculations. It is known that $x^n - 1$ may be factored out as $(x-1)(1+x+x^2+\cdots+x^{n-1})$. Using that fact consecutively for all the brackets one may obtain:
$$
k\ \text{times}\begin{cases}
x^n-1 = (x-1)(1+x+\cdots+x^{n-1})\\
x^{n-1}-1 = (x-1)(1+x+\cdots+x^{n-2})\\
\cdots\\
x^{n-k+1}-1 = (x-1)(1+x+\cdots+x^{n-k})
\end{cases}
$$
For the denominator:
$$
k\ \text{times}\begin{cases}
(x-1) = (x-1)\\
(x^2-1) = (x-1)(1+x)\\
\cdots \\
(x^k-1) = (x-1)(1+x+\cdots+x^{k-1})
\end{cases}
$$
So if we denote the expression under the limit as $f(x)$ we get:
$$
f(x) = \frac{(x-1)^k\prod\sum\cdots}{(x-1)^k\prod \sum\cdots}
$$
Now if we let $x\to1$ we get:
$$
\lim_{x\to1}f(x) = \frac{(n-1)(n-2)\cdots (n-k)}{1\cdot 2\cdot 3\cdots (k-1)} = \frac{(n-1)(n-2)\cdots (n-k)}{(k-1)!}
$$
But this doesn't match the keys section which has $n\choose k$ as an answer. I've checked several times but couldn't spot a mistake. Looks like I'm missing a $+1$ somewhere.

A quick way to evaluate the limit:
First consider
\begin{align}
\lim_{x \to 1} \frac{x^{n-j+1} -1}{x^j -1} &= \lim_{x \to 1} \frac{(n-j+1) x^{n-j}}{j \, x^{j-1}} = \frac{n-j+1}{j}
\end{align}
now,
\begin{align}
\lim_{x\to 1}\frac{(x^n -1)(x^{n-1}-1)\cdots(x^{n-k+1}-1)}{(x-1)(x^2-1)\cdots(x^k-1)} &= 
\lim_{x \to 1} \prod_{j=1}^{k} \frac{x^{n-j+1} -1}{x^j -1} \\
&= \prod_{j=1}^{k} \, \lim_{x \to 1} \frac{x^{n-j+1} -1}{x^j -1} \\
&= \prod_{j=1}^{k} \frac{n-j+1}{j} \\
&= \frac{(n)(n-1) \cdots (n-k+1)}{k!} = \frac{n!}{k! \, (n-k)!} \\
&= \binom{n}{k}.
\end{align}
