The $n^{th}$ root of the geometric mean of binomial coefficients. $\{{C_k^n}\}_{k=0}^n$ are binomial coefficients. $G_n$ is their geometrical mean. 
Prove 
$$\lim\limits_{n\to\infty}{G_n}^{1/n}=\sqrt{e}$$

In fact, we have
$$ \lim_{n\to\infty}\left[\prod_{k=0}^{n}\binom{n}{k}\right]^{1/n^2} = \exp\left(1+2\int_{0}^{1}x\log x\; dx\right) = \sqrt{e}.$$
This follows from the identity
$$\frac{1}{n^2}\log \left[\prod_{k=0}^{n}\binom{n}{k}\right] = 2\sum_{j=1}^{n}\frac{j}{n}\log\left(\frac{j}{n}\right)\frac{1}{n} + \left(1+\frac{1}{n}\right)\log n - \left(1+\frac{2}{n}\right)\frac{1}{n}\log (n!),$$
together with the Stirling's formula.
In fact, I tried to write down the detailed derivation of this identity, but soon gave up since it's painstrikingly demanding to type $\LaTeX$ formulas in iPad2!
But you may begin with the identity
$$\log\binom{n}{k} = \log n! - \log k! - \log (n-k)!$$
and
$$ \log k! = \sum_{j=1}^{k} \log j,$$
and then you can change the order of summation.
