Limit of Binomial Product

Here is my solution.

We have the following identity (can be proved by induction, or just tracking the terms)

$\prod_{r=1}^{n} \binom{n}{r}=\prod_{r=1}^{n}r^{2r-n-1}$ Note that $\ln \prod_{r=1}^{n} \binom{n}{r}= 2\sum_{r=1}^{n} r \ln r - (n+1) \sum_{r=1}^{n} \ln r$

Now, note that by integration by parts $\sum_{k=1}^n k \ln k\sim \int^n_0 x \log x dx= \frac{1}{2}n^2 \ln n - \frac{n^2}{4}$

Using this and Stirling's approximation that $\ln n! \sim n\ln n - n$ we find $2\sum_{r=1}^{n} r \ln r - (n+1) \sum_{r=1}^{n} \ln r \approx n^2 \ln n - \frac{1}{2}n^2 - (n^2+n) \ln n +n^2+n=\frac{n^2}{2}-n\ln n +n$

So $\lim_{n \to \infty}\frac{ \ln \prod_{r=1}^{n} \binom{n}{r}}{n^2+n}= \lim_{n \to \infty} \frac{n^2-2n \ln n+2n}{2n^2+2n} =\frac{1}{2}$

So the desired limit is

$\lim_{n\rightarrow \infty} \sqrt[n^2 + n ]{ \prod { n \choose r } } = e ^ { \displaystyle \lim_{n \to \infty}\frac{ \ln \prod_{r=1}^{n} \binom{n}{r}}{n^2+n} } = e ^ { \frac{1}{2}}.$

Note that $\prod_{j=0}^n \binom{n}{j} \; = \; \frac{(n!)^{n+1}}{\left(\prod_{j=0}^n j!\right)^2} \; =\; \frac{(n!)^{n+1}}{(n\$)^2} \; = \; \frac{(n!)^{n+1}}{G(n+2)^2}$ where $n\$ = \prod_{j=1}^n j!$ is what Plouffe calls the superfactorial, and $G$ is the Barnes G-function. Thus we are interested in the limit $X \; = \; \lim_{N \to \infty} \frac{(n!)^{\frac{1}{n}}}{G(n+2)^{\frac{2}{n(n+1)}}}$ Using Stirlings' approximation $n! \; \sim \; \sqrt{2\pi} e^{-n} n^{n+\frac12}$ and a similar asymptotic result for the Barnes G-function $G(n+1) \; \sim \; \frac{1}{A}(2\pi)^{\frac12n} n^{\frac12n^2 - \frac{1}{12}} e^{-\frac34n^2 + \tfrac{1}{12}}$ as $n \to \infty$ , where $A$ is Glaisher's constant , we deduce that $(n!)^{\frac{1}{n}} \; \sim \; \frac{n}{e} \hspace{1cm} G(n+2)^{\frac{1}{n(n+1)}} \; \sim \; \frac{\sqrt{n}}{e^{\frac34}} \hspace{2cm} n \to \infty$ and hence that $X = \sqrt{e} = \boxed{1.648721271}$ .