No simple duplication formula for factorials? Many special functions including the gamma function have a duplication formula of some sorts.  In the case of the gamma function it reads:

Gamma(2z) = Gamma(z) Gamma(z+1/2) 22z-1/Gamma(1/2)

On the other hand, there is no algebraic relation between Gamma(2z) and Gamma(z) by themselves, meaning that is there is no nonzero polynomial f(x,y) such that f(Gamma(2z),Gamma(z))=0 for all complex z.  I can prove this by chasing poles and their order.
However, I'd be interested in a (simple) argument which shows that the following similar statement is true (which I believe it is):

There is no (nonzero) polynomial f(x,y) such that f((2n)!, n!)=0 for all integers n≥0.

Any ideas?  Thank you!
 A: Armin,
Let me try to solve your original problem differently.
First write the wanted polynomial in the form $f(x,y)=\sum_kx^kA_k(y)$ where
the leading polynomial $A_0(y)$ is not identically zero (otherwise we can
always replace $f(x,y)$ by $f(x,y)/x^\ell$ for a suitable $\ell$). Denote by $N$
the degree of the polynomial $A_0(y)$. For any prime $p>N!$ the numbers
$0$ and $(-1)^kk!$, where $k=0,\dots,N-1$, are distinct residues modulo $p$, 
so that $p!\equiv 0\pmod p$ and $(p-k)!=(p-1)!/\prod_{j=1}^{k-1}(p-j)\equiv(-1)^k(k-1)!^{-1}\pmod p$
are pairwise noncongruent modulo $p$ as well. Substituting $x=(2p-2k)!\equiv0\pmod p$ 
and $y=(p-k)!$ for each $k=0,1,\dots,N$ into $f(x,y)=0$ and reducing modulo $p$, we obtain
$N+1$ different solutions of the polynomial equation 
$A_0(x)\equiv0\pmod p$, so that all coefficients of $A_0(x)$ are divisible by $p$.
Since this is true for any prime $p>N!$, the polynomial $A_0(x)$ is identically zero,
which contradicts our assumption.
Is it elementary enough?
A: I don't know how to answer your question, but here is something related that you will find interesting. Dick Lipton has a post about how a duplication formula that directly related Γ(2z) with Γ(z) would lead to an efficient classical algorithm for factoring.  If I remember correctly, he also speculates about potential approaches to showing that such formulae can't exist. 
A: It's equivalent to show that there is no polynomial relationship f({2n choose n}, n!) = 0.  On the other hand, we know that {2n choose n} ~ 4^n/sqrt{n} asymptotically and n! grows much faster.  
Terence Tao once remarked that if a sufficiently simple duplication formula were known for the factorial then Wilson's theorem would give an efficient primality test.  (Edit:  see the other answer.  I may be misremembering the stronger remarks that Dick Lipton made.)
