calculus - 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
$$\lim\limits_{n\to\infty}{G_n}^{1/n}=\sqrt{e}$$

Answer

$G_n$ is the geometric mean of $n+1$ numbers:
$$
G_n=\left[\prod_{k=0}^n{n\choose k}\right]^{\frac1{n+1}}
$$
or with $\log$ representing the natural logarithm (to the base $e$),
$$
\log G_n
= \frac1{n+1} \sum_{k=0}^n \log {n\choose k}
= \log n! - \frac2{n+1} \sum_{k=0}^n \log k!

\,.
$$
Stirling's approximation is
$n! \approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$
or
$$
\log n!
\approx
\frac12\log{(2\pi n)}+n\log\left(\frac{n}{e}\right)
= \left(n+\frac12\right)\log n+\frac12\log 2\pi-n

$$
so
$$
\eqalign{
\log \left(G_n\right)^\frac1n
&
= \frac1n \log G_n
= \frac1n \log n! - \frac2{n(n+1)} \log \prod_{k=0}^n k!
\\
&

= \frac1n \log n! - \frac2{n(n+1)} \sum_{k=0}^n \log k!
\\
&
\approx \left(1+\frac1{2n}\right) \log n
- \frac2{n(n+1)} \sum_{k=1}^n \left(k+\frac12\right)\log k
- \frac1{2n}\log 2\pi
\\
&
\approx \left(1+\frac1{2n}\right) \log n
- \frac2{n(n+1)}

\left[
\frac{n(n+1)}{2}\log n -
\frac{n(n+2)}{4}
\right]
- \frac1{2n}\log 2\pi
\\
&
= \frac{\log n-\log 2\pi}{2n}
+ \frac{n+2}{2(n+1)}
\\

&
\rightarrow \frac12
\,,
}
$$
where the sum of logarithms was approximated
using the definite integrals
$$
\sum_{k=1}^n \log k \approx
\int_1^n \log x\,dx =

\Big[x\log x-x\Big]_1^n \approx
\Big[x\log x-x\Big]_0^n
$$
and
$$
\sum_{k=1}^n k \log k \approx
\int_0^n x\log x\,dx=\left[\frac{x^2}{2}\log x - \frac{x^2}{4}\right]_0^n
$$
(using integration by parts as shown in a comment), so that
$$

\eqalign{
\sum_{k=1}^n \left(k+\frac12\right)\log k
&=
\sum_{k=1}^n k \log k + \frac12
\sum_{k=1}^n \log k
\\
&\approx
\left( \frac{n^2}{2}\log n - \frac{n^2}{4} \right) + \frac12
\Big( n \log n - n \Big)
\\

&=
\frac{n^2+n}{2}\log n - \frac{n^2+2n}{4}
\,.
}
$$
Thus
$$
G_n=e^{\log G_n}\rightarrow e^{\frac12}=\sqrt{e}
\,.
$$