<s>[INST] Rational fraction expression for triangular powers of 2 Why does the following pattern hold?
$$1=1=2^0$$
$$\frac{2(2^2-1)}{3}=2=2^1$$
$$\frac{3^2(3^2-1)(3^2-2^2)}{3^2 \times 5}=8=2^3$$
$$\frac{4^2(4^2-1)^2(4^2-2^2)(4^2-3^2)}{3^3 \times 5^2\times 7}=64=2^6$$
$$\frac{5^3(5^2-1)^2(5^2-2^2)^2(5^2-3^2)(5^2-4^2)}{3^4 \times 5^3 \times 7^2 \times 9}=1024=2^{10}$$
$$ n^{\lceil n/2\rceil}\prod_{k=1}^{n-1}\frac {(n^2-k^2)^{\lfloor (n-k+1)/2\rfloor}}{(2k+1)^{n-k}}=2^{n(n-1)/2}.$$
The motivation for the question comes from the following two questions:
Coefficients of binomial continued fractions
Determinants of products of binary matrices and binomial coefficients
 [/INST] a partial answer:
$$
\frac{5^3(5^2-1)^2(5^2-2^2)^2(5^2-3^2)(5^2-4^2)}{3^4 \times 5^3 \times 7^2 \times 9}
$$
may be re-written as
$$
\frac{\frac{9!}{0!}\frac{7!}{2!}\frac{5!}{4!}}{\frac{9!}{4!2^4}\frac{7!}{3!2^3}\frac{5!}{2!2^2}\frac{3!}{1!2^1}}
$$
</s>