Calculation of $ \lim_{x\rightarrow 1}\frac{(1-x)\cdot(1-x^2)\cdot(1-x^3)\cdots(1-x^{2n})}{\{(1-x)\cdot(1-x^2)\cdot (1-x^3)\cdots(1-x^n)\}^2} = $ Calculation of $\displaystyle \lim_{x\rightarrow 1}\frac{(1-x)\cdot(1-x^2)\cdot(1-x^3)\cdots (1-x^{2n})}{\{(1-x)\cdot(1-x^2)\cdot (1-x^3)\cdots(1-x^n)\}^2} = $
My Trial After simplification, we get $$\displaystyle \lim_{x\rightarrow 1}\frac{(1-x^{n+1})\cdot(1-x^{n+2})\cdot(1-x^{n+3})\cdots(1-x^{2n})}{(1-x)\cdot(1-x^2)\cdot (1-x^3)\cdots(1-x^n)}$$
Now I did not understand How can I solve after that, Help me
Thanks
 A: By L'Hospital's rule, 
\begin{align*}
\lim_{x \to 1} \frac{1 - x^{n + k}}{1 - x^k} &= \lim_{x \to 1} \frac{-(n + k) x^{n + k}}{-k x^k} = \frac{n + k}{k}
\end{align*}
Hence the desired limit is
$$\frac{n + 1}{1} \cdot \frac{n + 2}{2} \cdot \frac{n + 3}{3} \cdots \frac{n + n}{n} = \frac{(2n)!}{(n!)^2}$$
Alternatively, factor out a common term of $1 - x$ to find
$$\frac{1 - x^{n + k}}{1 - x^k} = \frac{1 + x + x^2 + \dots + x^{n + k - 1}}{1 + x + x^2 + \dots + x^{k - 1}}$$
Now directly evaluate at $1$ to get $n + k$ in the numerator and $k$ in the denominator.
A: Try the factorization $(1-x^n) = (1-x)(1+x+\dots+x^{n-1})$.
A: We can use the identity: $$1 - x^m = (1-x)(1 + x + \cdots+ x^{m-1})$$
in every single factor, there. All of those $1-x$ will cancel, and we'll be left with $$\lim_{x \to 1} \frac{(1+x+\cdots+ x^n)(1+x+\cdots+ x^{n+1})\cdots (1+x+\cdots +x^{2n - 1})}{(1+x)(1+x+x^2)\cdots(1+x+x^2+\cdots +x^{n-1})} = \frac{(n+1)(n+2)\cdots 2n}{1\cdot 2 \cdot 3 \cdots n}$$
To put this nicely, see that $$\frac{(n+1)(n+2)\cdots 2n}{1\cdot 2 \cdot 3 \cdots n} = \frac{1\cdot 2 \cdot 3 \cdots n \cdot (n+1)\cdot (n+2)\cdots 2n}{ 1\cdot 2 \cdot 3 \cdots n \cdot 1\cdot 2 \cdot 3 \cdots n} = \frac{(2n)!}{(n!)^2}$$
