Finding a limit where binomial coefficients appear as powers. This one is for my mate that is 2 years older than me. Could you help please?
$$\lim_{n\to\infty}\frac{\sqrt[\uproot{3}\Large 2^n]{n^{\textstyle\binom{n}{0}}\cdot (n+1)^{\textstyle\binom{n}{1\vphantom{0}}}\cdots(n+n)^{\textstyle\binom{n}{n\vphantom{0}}} }}{n}$$
 A: Hint: logarithm turns a product into a sum.
A: Ok, I think I have a way to show that the limit is $\frac{3}{2}$.
If you take logs, you get
$$x_n  = \frac{1}{2^n} \sum_{r=0}^{n} \left(\binom{n}{r} (\log(n+r)-\log n)\right)  = \frac{1}{2^n} \sum_{r=0}^{n} \binom{n}{r} \log\left(1+\frac{r}{n}\right)$$
Now we use the fact that, if $f$ is continuous on $[0,1]$, then 
$$f(t) = \lim_{n \to \infty} \sum_{r=0}^{n} \left(\binom{n}{r} t^r(1-t)^{n-r}f\left(\frac{r}{n}\right)\right) $$
For a proof of this fact, and for more information, see: Bernstein Polynomials. Bernstein polynomials can be used to give a proof of the Weierstrass approximation theorem, and also have probabilistic interpretations (I am guessing Prof Israel was probably referring to something equivalent).
In this case, we basically have $f(t) = \log (1+t)$ and $t=\frac{1}{2}$.
Thus $x_n \to \log(\frac{3}{2})$ and your limit is $\frac{3}{2}$.
A: It has been almost 3 days. I hope it is okay to provide an almost complete answer.
$$
\begin{align}
&\log\left(\lim_{n\to\infty}\frac{\left((n+0)^{\binom{n}{0}}(n+1)^{\binom{n}{1}}(n+2)^{\binom{n}{2}}\dots(n+n)^{\binom{n}{n}}\right)^{1/2^n}}{n}\right)\\
&=\lim_{n\to\infty}\frac1{2^n}\sum_{k=0}^n\binom{n}{k}\log\left(1+\frac kn\right)\tag{1}
\end{align}
$$

One approach to handle $(1)$ uses a simply proven version of the law of large numbers for binomial coefficients
Since $t^{\alpha n}\le t^k$ for $k\le\alpha n$ and $t\in[0,1]$, we have the bound
$$
\begin{align}
t^{\alpha n}\sum_{k/n\in[0,\alpha]}\binom{n}{k}\lambda^k(1-\lambda)^{n-k}
&\le\sum_{k=0}^n\binom{n}{k}\lambda^k(1-\lambda)^{n-k}t^k\\
&=(\lambda t+1-\lambda)^n\tag{2}
\end{align}
$$
Therefore,
$$
\begin{align}
\sum_{k/n\in[0,\alpha]}\binom{n}{k}\lambda^k(1-\lambda)^{n-k}
&\le\left(\frac{\lambda t+1-\lambda}{t^\alpha}\right)^n\\
&=\left(\frac{\lambda^\alpha(1-\lambda)^{1-\alpha}}{\alpha^\alpha(1-\alpha)^{1-\alpha}}\right)^n\tag{3}
\end{align}
$$
when $t=\frac{(1-\lambda)\alpha}{\lambda(1-\alpha)}$. Note that to ensure $t\in[0,1]$, we need $\alpha\in[0,\lambda]$.
As $\alpha$ increases from $0$ to $\lambda$, $\dfrac{\lambda^\alpha(1-\lambda)^{1-\alpha}}{\alpha^\alpha(1-\alpha)^{1-\alpha}}$ increases from $1-\lambda$ to $1$.
Thus, for any $\alpha\lt\lambda\lt\beta$, $(3)$ and symmetry yields
$$
\lim_{n\to\infty}\sum_{k/n\not\in(\alpha,\beta)}\binom{n}{k}\lambda^k(1-\lambda)^{n-k}=0\tag{4}
$$

Thus, for any $\epsilon\gt0$, we can find an $\alpha\lt1/2$ so that
$$
k/n\in(\alpha,1-\alpha)\Rightarrow|\log(1+k/n)-\log(3/2)|\lt\epsilon/2\tag{5}
$$
Then, using $(4)$ with $\lambda=1/2$ and $\beta=1-\alpha$, we can find an $N$ so that
$$
n\ge N\Rightarrow\frac1{2^n}\sum_{k/n\not\in(\alpha,1-\alpha)}\binom{n}{k}\le\epsilon/2\tag{6}
$$
Then
$$
\begin{align}
&\left|\frac1{2^n}\sum_{k=0}^n\binom{n}{k}\log\left(1+\frac kn\right)\ -\ \log\left(\frac32\right)\right|\\[6pt]
&=\left|\frac1{2^n}\sum_{k=0}^n\binom{n}{k}\left[\log\left(1+\frac kn\right)-\log\left(\frac32\right)\right]\right|\\[6pt]
&\le\left|\frac1{2^n}\sum_{k/n\in(\alpha,1-\alpha)}\binom{n}{k}\left[\log\left(1+\frac kn\right)-\log\left(\frac32\right)\right]\right|\tag*{use $(5)$}\\
&+\left|\frac1{2^n}\sum_{k/n\not\in(\alpha,1-\alpha)}\binom{n}{k}\left[\log\left(1+\frac kn\right)-\log\left(\frac32\right)\right]\right|\tag*{use $(6)$}\\[6pt]
&\le\epsilon/2+\epsilon/2\\[6pt]
&=\epsilon\tag{7}
\end{align}
$$
Since $\epsilon\gt0$ was arbitrary, we have
$$
\lim_{n\to\infty}\frac1{2^n}\sum_{k=0}^n\binom{n}{k}\log\left(1+\frac kn\right)=\log\left(\frac32\right)\tag{8}
$$
Plugging $(8)$ back into $(1)$, we get the final answer.
