Find $\lim\limits_{n\to+\infty}\sqrt[n]{\prod\limits_{k=1}^{n}{n\choose k}}$ I tried to compute the product of binomial coefficients. I found that 
$$\prod\limits_{k=0}^{n}{n\choose k}=\frac{H^2(n)}{(n!)^{n+1}}$$
I am not familiar with hyperfactorial function. How to find this binomial product? 
 A: Take logarithm of the limit, because the logarithm is a continuous function we have
$$\log L := \lim_{n\to \infty} \frac{1}{n} \sum_{k=1}^{n} \log \begin{pmatrix} n \\k\end{pmatrix}.$$
Now call the sequence in the numerator $a_n$ and the sequence in the denominator $b_n$ and use Cesàro-Stolz theorem, i.e.
$$L= \lim_{n\to \infty} \frac{a_n}{b_n} = \lim_{n\to \infty} \frac{a_{n+1}-a_n}{b_{n+1}-b_n}.$$
(Check that the hypotheses for this theorem are fulfilled)
So,
$$a_{n+1}- a_n = \sum_{k=1}^n \left[\log \begin{pmatrix} n+1 \\k\end{pmatrix}- \log \begin{pmatrix} n \\k\end{pmatrix} \right] = \sum_{k=1}^n \log \frac{n+1}{n+1-k}.$$
Hence,
$$\log L = \lim_{n\to \infty} \sum_{k=1}^n \log \frac{n+1}{n+1-k}=\lim_{n\to \infty} \left[ n \log (n+1) - \sum_{k=1}^n \log k\right].$$
From this one should deduce $\log L=\infty$ and hence $L=\infty$.
A: An heuristic (not well justified, though in agreement with numerical evaluation):
Using the well know approximation $$\log {n \choose k}\approx n \, h\left(\frac{k}{n}\right)$$
where $h(x)=-x \log(x) - (1-x) \log(1-x)$  (entropy function, in nats), we can write
$$\log \prod\limits_{k=0}^{n}{n\choose k} =\sum_{k=0}^{n} \log {n\choose k}\approx \int_0^n n \, h\left(\frac{k}{n}\right) dk = n^2 \int_0^1h(x) dx = \frac{n^2}{2}$$
(numerical evaluation suggests that the next term is $- n \log(n)/2$ )
Then, $$\sqrt[n]{\prod\limits_{k=1}^{n}{n\choose k}} \approx \sqrt[n]{\exp{ \left(\frac{n^2}{2} \right)}} = \exp{ \left(\frac{n}{2} \right)}$$
or, perhaps more interesting: 
$$\sqrt[n^2]{\prod\limits_{k=1}^{n}{n\choose k}}\to e^{1/2}=1.6487\cdots$$
Adding the following term: $$\sqrt[n^2]{\prod\limits_{k=1}^{n}{n\choose k}}\approx \exp\left[\frac{1}{2}\left(1- \frac{\log (n)}{n}\right)\right]=\sqrt{\frac{e}{n^{1/n}}}$$
A: If you are asking about the limit as $n\rightarrow\infty$ as in your title. 
Use 
$$\binom nk \geq \binom n1 =n \ \ \ \textrm{
for $1\leq k \leq n-1$.}$$
A: Considering $$A_m=\prod\limits_{k=0}^{m}{m\choose k}=\frac{H^2(m)}{(m!)^{m+1}}$$ $$\log(A_m)=2\log(H(m))-(m+1)\log(m!)$$ Now, refering to this question $$\log(H(m))=m^2 \left(\frac{1}{2} \log(m)-\frac{1}{4}\right)+\frac{1}{2} m
   \log (m)+\left(\log (A)+\frac{1}{12} \log
   (m)\right)+O\left(\left(\frac{1}{m}\right)^2\right)$$ and using Stirling approximation for the factorial leads to $$\log(A_m)=\frac{m^2}{2}+\frac{1}{2} m \left(2-\log (m)-\log (2 \pi
   )\right)+\frac{1}{12} \left(24 \log (A)-4 \log (m)-1-6 \log
   (2 \pi )\right)+O\left(\frac{1}{m}\right)$$ So $$\frac 1 m \log(A_m) \approx \frac m2+\cdots$$ and $$\Big(A_m\Big)^{\frac 1m}\approx e^{\frac m2}$$ or, much more accurate, $$\Big(A_m\Big)^{\frac 1m}\approx\frac{e^{\frac{m}{2}+1}}{\sqrt{2 \pi m} }$$
