Finding: $\lim\limits_{m\to\infty}\left(\sqrt[m]{m+1}\cdot\sqrt[m^2+m]{\binom{m}{1}\binom{m}{2}\cdots\binom{m}{m}}\right) .$ 
I am having trouble with the limit;
   $$\lim_{m\to\infty}\left(\sqrt[m]{m+1}\cdot\sqrt[m^2+m]{\binom{m}{1}\binom{m}{2}\cdots\binom{m}{m}}\right) .$$

where $$ \binom{m}{j}=\frac{m!}{(m-j)!j!}$$
I tried to use the Stirling formula but still I could not clarify the steps.
By the way Mathematica says the answer should be $\sqrt{e}$.
Any nice trick for this ?
 A: Note that
$$\sqrt[m]{m+1}=e^{\frac{\log(m+1)}{m}}\to e^0=1$$
and
$$\sqrt[m^2+m]{\binom{m}{1}\binom{m}{2}\cdots\binom{m}{m}}=e^{\frac{\log\left(\sum_{k=1}^m\binom{m}{k}\right)}{m^2+m}}\to\sqrt e$$
indeed by Stolz-Cesaro:
$$\lim_{m\to\infty}\frac{\sum_{k=1}^m\log\binom{m}{k}}{m^2+m}=\lim_{m\to\infty}\frac{\sum_{k=1}^{m+1}\log\binom{m+1}{k}-\sum_{k=1}^m\log\binom{m}{k}}{(m+1)^2+(m+1)-(m^2+m)}=\lim_{m\to\infty}\frac{\sum_{k=1}^m\log \frac{m+1}{m+1-k}}{2m+2}=\lim_{m\to\infty}\frac{\log \frac{(m+1)^m}{m!}}{2m+2}=\lim_{m\to\infty}\frac12\cdot \frac{m}{m+1}\cdot \log \frac{(m+1)}{\sqrt[m]{m!}}=\frac12\cdot 1 \cdot \log e=\frac12$$
A: Using the technique of double summation,
$$ \sum_{k=1}^{n} \log (k!) = \sum_{k=1}^{n} \sum_{j=1}^{k} \log j = \sum_{j=1}^{n} (n+1-j) \log j. $$
So it follows that
\begin{align*}
\sum_{k=1}^{m} \log \binom{m}{k}
&= (m+1) \log (m!) - 2 \sum_{k=1}^{m} \log (k!) \\
&= \sum_{j=1}^{m} (2j-m-1) \log j \\
&= \sum_{j=1}^{m} (2j-m-1) \log \left(\frac{j}{m}\right)
\end{align*}
The last line follows from the fact that $\sum_{j=1}^{m} (2j-m-1) \log m = 0$. So we have
\begin{align*}
\frac{1}{m(m+1)}\sum_{k=1}^{m} \log \binom{m}{k}
&= \frac{m}{m+1} \sum_{j=1}^{m} \left(\frac{2j-1}{m}-1\right) \log \left(\frac{j}{m}\right) \, \frac{1}{m} \\
&\xrightarrow[m\to\infty]{} \int_{0}^{1} (2x-1)\log x \, dx = \frac{1}{2}.
\end{align*}
(Although justifying this convergence requires more than the usual Riemann sum argument, it is still manageable.)
Together with $(m+1)^{1/m} \to 1$, we conclude that the limit is $e^{1/2} = \sqrt{e}$.
