Computing: $\lim\limits_{n\to\infty}\left(\prod\limits_{k=1}^{n} \binom{n}{k}\right)^\frac{1}{n}$ I try to compute the following limit:
$$\lim_{n\to\infty}\left(\prod_{k=1}^{n} \binom{n}{k}\right)^\frac{1}{n}$$
I'm interested in finding some reasonable ways of solving the limit. I don't find any easy approach. Any hint/suggestion is very welcome.
 A: Start with the identity $$\prod_{j=1}^n {n\choose j} =\prod_{k=1}^n {k^k\over k!}.$$
Since $k^k/k!\to \infty$, the same is true of its geometric mean: 
$$\left(\prod_{k=1}^n {k^k\over k!}\right)^{1/n}\to\infty.$$ 
A: As noted by Phira and Byron Schmuland, the product diverges. 
I find an asymptotic expression for the product for large $n$, Eqn. (3) below.
I have found some inspiration for this solution in @sos440's answer here. 
With a little work, one can show that 
$$\begin{equation*}
\log \left(\prod_{k=1}^n{n\choose k}\right)^{1/n} 
= -\frac{n+1}{n}\log n! 
+ (n+1)\log n
+ 2 \sum_{j=1}^n \frac{j}{n}\log\frac{j}{n}. \tag{1}
\end{equation*}$$
For a derivation of (1), see below. 
Using Stirling's approximation, and the fact that 
$\sum_{j=1}^n \frac{j}{n}\log\frac{j}{n} \approx n\int_0^1 x \log x = -n/4$ 
(the error here is $O(\log(n)/n)$),
we get
$$\begin{equation*}
\log \left(\prod_{k=1}^n{n\choose k}\right)^{1/n}
\sim \frac{n}{2}+1 - \frac{1}{2} \log 2\pi n.\tag{2}
\end{equation*}$$
Therefore,
$$\begin{equation*}
\left(\prod_{k=1}^n{n\choose k}\right)^{1/n} \sim \frac{e^{n/2+1}}{\sqrt{2\pi n}}. \tag{3}
\end{equation*}$$
Clearly the product diverges.
For $n=10$, $100$, and $1000$ the left and right side of (3) agree to $12\%$, $2.0\%$, and $0.28\%$, respectively. 
From (3) we get the result 
$\lim_{n\to\infty} \left(\prod_{k=1}^n{n\choose k}\right)^{1/n^2} = \sqrt{e}$
for free.
(Use $\lim_{n\to\infty} x^{1/n} = 1$ for $0<x<1$.)
Derivation of Eqn. (1)
Note that 
$$\begin{eqnarray*}
\log \left(\prod_{k=1}^n{n\choose k}\right)^{1/n}
&=& \log \left(\prod_{k=0}^n{n\choose k}\right)^{1/n} \\
&=& \frac{1}{n} \sum_{k=0}^n \log {n\choose k} \\
&=& \frac{1}{n} \sum_{k=0}^n \left(\log n! - \log k! - \log (n-k)! \right)\\
&=& \frac{1}{n} \left((n+1)\log n!  - 2 \sum_{k=0}^n\log k!  \right).
\end{eqnarray*}$$
But 
$$\begin{eqnarray*}
\sum_{k=0}^n\log k! &=& \sum_{k=1}^n\log k! \\
&=& \sum_{k=1}^n \sum_{j=1}^k \log j \\
&=& \sum_{j=1}^n \sum_{k=j}^n \log j \\
&=& \sum_{j=1}^n (n+1-j) \log j \\
&=& (n+1)\sum_{j=1}^n \log j - \sum_{j=1}^n j (\log j -\log n + \log n) \\
&=& (n+1)\log n! - \frac{n(n+1)}{2} \log n - \sum_{j=1}^n j\log\frac{j}{n}.
\end{eqnarray*}$$
Eqn. (1) follows immediately. 
Some key identities: 
$$\begin{eqnarray*}
\log n! &=& \sum_{k=1}^n \log k \\
\sum_{k=j}^n 1 &=& n+1-j \\
\sum_{k=1}^n k &=& \frac{n(n+1)}{2} \\
\sum_{k=1}^n \sum_{j=1}^k a_{j k} &=& \sum_{j=1}^n \sum_{k=j}^n a_{j k}
\end{eqnarray*}$$
A: All the binomial coefficients except the last one are at least $n$, so the $n$th root is at least $n^{\frac{n-1}{n}}$, so the limit is infinity.
