Limit of Stirling approximation theorem This limit I tried to solve it by using Stirling approximation and here is the limit and my tried 
$\lim_{0\to\infty }\frac{ \sqrt[n]{\dbinom{n}{1} \dbinom{n}{2} \dbinom{n}{3} \dbinom{n}{4} \cdots \dbinom{n}{n} } }{e^{\frac{n}{2}} n^{\frac{-1}{2}}}=? $ ..... I tried to evaluate the numerator by Stirling approximation and the denominator and got that 
$\lim \limits_{n\to \infty }\frac{\sqrt{2\pi n} e^{\frac{-3}{2} n}n^{\frac{2n+1}{2} }}{\left[ 1! 2! 3! 4! \cdots n! \right] \left[ \left( n-1\right) ! \left( n-2\right) ! \left( n-3\right) ! \cdots 0! \right] }$...  My problem that I can't evaluate the denominator..  what should I do next to solve this limit?.  Any hint?  Thanks.
 A: Using the Euler-Maclaurin Sum Formula, we get
$$
\frac2n\sum_{k=1}^n(k\log(k)-k)\sim n\log(n)-\frac32n+\log(n)-1\tag{1}
$$
Since $\binom{n}{k}=\frac{n!}{k!\,(n-k)!}$, it follows that $\prod\limits_{k=0}^n\binom{n}{k}=\frac{n!^{n+1}}{\color{#C0C0C0}{0!}1!2!\cdots n!\cdot n!(n-1)!\cdots1!\color{#C0C0C0}{0!}}=\frac{n!^{n+1}}{\left(\prod\limits_{k=1}^nk!\right)^2}$
Therefore,
$$
\begin{align}
\left[\prod_{k=0}^n\binom{n}{k}\right]^{1/n}
&=\frac{n!^{\frac{n+1}n}}{\left(\prod\limits_{k=1}^nk!\right)^{\frac2n}}\tag{2}\\
&\sim\frac{n!^{\frac{n+1}n}}{\left(\prod\limits_{k=1}^n\sqrt{2\pi k}k^ke^{-k}\right)^{\frac2n}}\tag{3}\\
&=\frac{n!}{2\pi\left(\prod\limits_{k=1}^nk^ke^{-k}\right)^{\frac2n}}\tag{4}\\
&\sim\frac{n!}{2\pi n^{n+1}e^{-\frac32n}e^{-1}}\tag{5}\\[12pt]
&\sim\bbox[5px,border:2px solid #C0A000]{\frac{e^{\frac n2+1}}{\sqrt{2\pi n}}}\tag{6}
\end{align}
$$
Explanation:
$(2)$: rearrange terms
$(3)$: use Stirling to approximate $k!$
$(4)$: $\sqrt{2\pi k}$ comes out of the product to the $\frac2n$ power as $2\pi n!^{\frac1n}$
$(5)$: apply $(1)$
$(6)$: use Stirling to approximate $n!$
This seems to be good:
at $n=1000$, the left side is $4.79983\times10^{215}$ and the right is $4.81333\times10^{215}$
at $n=10000$, the left side is $3.217057\times10^{2169}$ and the right is $3.218208\times10^{2169}$.
