Closed form for $\prod_{k=0}^n\binom{n}{k}x^ky^{n-k}$.

Question

This is well known that we have :

$$\forall (x,y) \in \mathbb{C}^2, ~ \forall n \in \mathbb{N}^*, ~ (x+y)^n=\sum_{k=0}^n\binom{n}{k}x^ky^{n-k}$$

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Now I don't know if there is any possible application but in same conditions (at least for real numbers) is there a close form to this one?

$$\prod_{k=0}^n\binom{n}{k}x^ky^{n-k}$$

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For real numbers I think in all cases (but I am not sure) if there is a close the form may look like $A\times x^a y^b$ where $A$ is an integer from the product of the binomial terms.

I looked at some values on Wolfram-Alpha with $(x,1-x)$ and $|x|<1$ for fun but I have some difficulties to find a close form a least to $A$.

I would be gald if someone has an idea.

Answer 1

With a bit of algebra $$\prod_{k=0}^n\binom{n}{k}=\frac{(n!)^{n+1}}{\prod_{k=0}^n(k!)^2}=\frac{(n!)^{n+1}}{(\prod_{k=1}^nk^{n-k+1})^2}=\prod_{k=1}^nk^{-n+2k-1}=\frac1{(n!)^{n+1}}\left(\color{red}{\prod_{k=1}^nk^k}\right)^2$$

where the expression in red doesnt seems easy to simplify. Of course as @Thomas showed in the comment$$\prod_{k=0}^n x^ky^{n-k}=(xy)^{\binom{n+1}{2}}$$

EDIT: it seems that we can get an analytic form (but not closed due to the periodicity of the $\tilde B_k$ functions) for the red term using the Euler-Maclaurin formula, something like

$$\left(\prod_{k=1}^nk^k\right)^2=\exp\left(\sum_{k=1}^n 2k\ln k\right)=\exp\left(\int_1^n \left(2t\ln t+\frac{\tilde B_3(t)}{3t^2}\right)\mathrm dt+ \left(n+\frac16\right)\ln n\right)$$

where $\tilde B_3$ is the $1$-periodic extension of the Bernoulli polynomial on $[0,1]$

$$B_3(x)=x^3-\frac32x^2+\frac12x$$

Answer 2

There is no "closed" form for this expression, but we can reduce it to another form: $$\prod_{k=0}^n\binom{n}{k}x^ky^{n-k} = (xy)^{n(n+1)/2} \prod_{k=0}^n \dfrac{n!}{k!(n-k)!} = (xy)^{n(n+1)/2} (n!)^{n+1} \prod_{k=0}^n \dfrac{1}{k!(n-k)!} = \\= (xy)^{n(n+1)/2} (n!)^{n+1}\prod_{k=0}^n\dfrac{1}{k!}\prod_{k=0}^n\dfrac{1}{(n-k)!} = (xy)^{n(n+1)/2} (n!)^{n+1}\left(\prod_{k=0}^n\dfrac{1}{k!}\right)^2 = \\ = (xy)^{n(n+1)/2} (n!)^{n+1}\dfrac{1}{1^2 \cdot 1^22^2 \cdot 1^22^23^2 \cdot \dots \cdot 1^22^2\dots n^2} = (xy)^{n(n+1)/2} (n!)^{n+1}\left(\prod_{k=1}^n\dfrac{1}{k^{n-k+1}}\right)^2$$

Answer 3

To complement Masacroso answer, note that the product of $k^k$ can be rewritten as $$ \eqalign{ & \prod\limits_{k = 1}^n {k^{\,k} } = \prod\limits_{k = 1}^n {\left( {{{\Gamma \left( {k + 1} \right)} \over {\Gamma \left( k \right)}}} \right)^{\,k} } = \cr & = {{\Gamma (2)^{\,1} } \over {\Gamma (1)^{\,1} }}{{\Gamma (3)^{\,2} } \over {\Gamma (2)^{\,2} }}{{\Gamma (4)^{\,3} } \over {\Gamma (3)^{\,3} }}\; \cdots \;{{\Gamma (n + 1)^{\,n} } \over {\Gamma (n)^{\,n} }} = \; \cr & = {{\Gamma (n + 1)^{\,n} } \over {\prod\limits_{1\, \le \,k\, \le \;n} {\Gamma (k)} }} = {{G(1)} \over {G(n + 1)}}\left( {n!} \right)^{\,n} \cr} $$

where $G(x)$ denotes the Barnes G_Function