Fast algorithm computing multiplications of 2$^{n}$ number of decreasing natural numbers Let A=n(n-1)(n-2)(n-3)  ... 4 multiplications.
But, we can reduce it to using 2 multiplications.

*

*n(n-3)= n$^{2}$-3n = X

*(n-1)(n-2)= n$^{2}$-3n+2 = X+2

Therefore: A=X*(X+2) ... 2 multiplications.
Because ((n+a)$^{2}$+b) is a polynomial of order 2.
(((n+a)$^{2}$+b)$^{2}$+c) is of order 4 using 2 multiplications.
((((n+a)$^{2}$+b)$^{2}$+c)$^{2}$+d) is of order 8 using 3 multiplications
...
In general, I guess  multiplication of 2$^{n}$ number of decreasing(or increasing) natural numbers can be done by just using n multiplications.
But how, if it is possible?
Conclusion:
miracle173 suggested that computing those unknown variables a,b,c,d,... in
the guess, general form of repeating squaring would be as difficult as factoring a number.
While Simon gave an algorithm that shrinks the problem to
twice faster than directly multiplying from the
problem definition.
Update1:
I just figured out the solution: n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)=
((n$^{2}$-7n+7)$^{2}$-21)$^{2}$+448n-784-64n$^{2}$
I.e. 2$^{3}$ number of decreasing natural numbers can be computed using just 3 multiplications, though not exactly the form in the beginning guess.
So it is reasonable to move on the conjecture to the next level: Could computing the product of 16 decreasing/increasing
natual numbers be done by using 4 multiplications?
 A: Expanding an idea mentioned here: https://codeforces.com/blog/entry/63491?#comment-474113
For the following considerations we assume that $m! \bmod n$ can be calculated in polynomial time. This is the case if $m!$ can be calculated in polynomial number $O(\log m)$  of multiplications and additions. Then you would be able to calculate $([\sqrt n  ]+1)! \bmod n$ in polynomial time. If $n$ is composite its smallest prime factor $p$ is smaller or equal $\sqrt n$ and so  $p | ([\sqrt n  ]+1)!$. Therefore  $p | ([\sqrt n  ]+1)! \bmod n$ and so $$p|\gcd(([\sqrt n  ]+1)! \bmod n,n).$$ Either you have found a proper divisor of $n$ or there is at least a divisor less or equal then $\sqrt[3]n$ and you can repeat this procedures. So if you continue in this way you would have a polynomial factorization algorithm. At the moment no polynomial factorization algorithm is known.
A: Indeed you found a nice trick to evaluate $n(n-1)(n-2)(n-3)$ with only 2 multiplications instead of the naive $3$. But Im afraid you are wrong about the generalization to larger products. Here is something that actually works:
Let $$f_k(n) = (n+2k-1)(n+2k-3)(n+2k-5)...(n-2k+5)(n-2k+3)(n-2k+1)$$
This has $2k$ terms in total, requiring $2k-1$ multiplications when done naively. But reordering the product gives
\begin{align}
f_k(n) &= (n+1)(n-1)(n+3)(n-3)(n+5)(n-5)...(n+2k-1)(n-2k+1) \\\\
&= (n^2-1)(n^2-3^2)(n^2-5^2)...(n^2-(2k-1)^2).
\end{align}
This way only takes $k$ multiplications (one to compute the square $n^2$ and $k-1$ to multiply all the terms). Your formula is a rescaled and shifted special case of this:
\begin{align}
n(n-1)(n-2)(n-3) &= f_2(2n-3)/16
\end{align}
similarly
\begin{align}
n(n-1)(n-2)(n-3)(n-4)(n-5) &= f_3(2n-5)/2^6 \\\\
n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7) &= f_4(2n-7)/2^8
\end{align}
and so on.
Overall this trick saves around a factor of two. Not more.
