In Apéry's proof of the irrationality of $\zeta(3)$, while proving the formula for the fast-converging series $\zeta(3)=\frac{5}{2} \sum_{k=1}^{\infty}{\frac{ (-1)^{k-1}} {k^{3}\binom {2k}{k}}}$ there is the following identity:
$$
\frac{(-1)^{n-1}(n-1)!^2}{n^2(n^2-1^2)\ldots(n^2-(n-1)^2)} = \frac{2(-1)^{n-1}}{n^2 \binom{2n}{n}},
$$
for positive integer $n>0$. I can't for the life of me figure out how to prove this is true. By eliminating on both sides the $(-1)^{n-1}$ in the numerator and the $n^2$ in the denominator, we are left with:
$$
\frac{(n-1)!^2}{(n^2-1^2)\ldots(n^2-(n-1)^2)} = \frac{2}{ \binom{2n}{n}}.
$$
But this already seems absurd! Using Wolfram Alpha shows that LHS-RHS is not zero, therefore the "identity" does not hold.
What am I missing? Note that I am using Van der Poorten's paper to understand Apéry's proof. The identity in question can be seen on the page labelled "197", on the bottom of the left column.
Best Answer
For each integer $k$, we have $n^2 - k^2 = \left(n-k\right) \left(n+k\right)$. Thus, \begin{align*} \prod\limits_{k=1}^{n-1} \left(n^2-k^2\right) &= \prod\limits_{k=1}^{n-1} \left(\left(n-k\right) \left(n+k\right)\right) = \underbrace{\left(\prod\limits_{k=1}^{n-1} \left(n-k\right) \right)}_{=\left(n-1\right)!} \underbrace{\left(\prod\limits_{k=1}^{n-1} \left(n+k\right)\right)}_{= \left(2n-1\right) \left(2n-2\right) \cdots \left(n+1\right)} \\ &= \left(n-1\right)! \left( \left(2n-1\right) \left(2n-2\right) \cdots \left(n+1\right) \right) . \end{align*} Hence, \begin{align*} \dfrac{\prod\limits_{k=1}^{n-1} \left(n^2-k^2\right)}{\left(n-1\right)!^2} &= \dfrac{\left(n-1\right)! \left( \left(2n-1\right) \left(2n-2\right) \cdots \left(n+1\right) \right)}{\left(n-1\right)!^2} \\ &= \dfrac{\left(2n-1\right) \left(2n-2\right) \cdots \left(n+1\right) }{\left(n-1\right)!} \\ &= \dfrac{\left(2n\right) \cdot \left( \left(2n-1\right) \left(2n-2\right) \cdots \left(n+1\right) \right)}{\left(2n\right) \cdot \left(n-1\right)!} \\ &= \dfrac{\left(2n\right) \left(2n-1\right) \left(2n-2\right) \cdots \left(n+1\right) }{2 n \cdot \left(n-1\right)!} \\ &= \dfrac{\left(2n\right) \left(2n-1\right) \left(2n-2\right) \cdots \left(n+1\right) }{2 n!} \qquad \left(\text{since } n \cdot \left(n-1\right)! = n! \right) \\ &= \dfrac{1}{2} \cdot \underbrace{\dfrac{\left(2n\right) \left(2n-1\right) \left(2n-2\right) \cdots \left(n+1\right) }{n!}}_{\substack{=\dbinom{2n}{n}\\ \text{(by the definition of }\dbinom{2n}{n}\text{)}}} \\ &= \dfrac{1}{2} \dbinom{2n}{n}. \end{align*} Taking the reciprocal of both sides, we obtain \begin{align*} \dfrac{\left(n-1\right)!^2}{\prod\limits_{k=1}^{n-1} \left(n^2-k^2\right)} = \dfrac{2}{\dbinom{2n}{n}} . \end{align*} In other words, \begin{align*} \dfrac{\left(n-1\right)!^2}{\left(n^2-1^2\right)\left(n^2-2^2\right)\cdots\left(n^2-\left(n-1\right)^2\right)} = \dfrac{2}{\dbinom{2n}{n}} . \end{align*}