Binomial coefficient identity found in Apéry's theorem 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.
 A: 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*}
