<s>[INST] 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.
 [/INST] You are indeed missing a $+1$ on the top. $x^2-1=(x-1)(x+1)$. You divide by $(x-1)$ and set $x=1$ and you get $2$, not $n-1=2-1=1$. In the $1+x+...+x^{n-1}$ there are $n$ terms, since you have $x^0$ as well. Same at the bottom, you go up to $k$, not $k-1$
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