Evaluating $\lim\limits_{n\to \infty}\left\{(1+\frac{1}{n})(1+\frac{2}{n})\dots(1+\frac{n}{n})\right\}^{\frac{1}{n}}$ Question is to evaluate :
$$\lim_{n\to \infty}\left\{ \left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\dots\left(1+\frac{n}{n}\right)\right\}^{\frac{1}{n}}$$
I tried to do something like this but it is not getting better...
$(1+\frac{1}{n})(1+\frac{2}{n})\cdots(1+\frac{n}{n})=(\frac{n+1}{n})(\frac{n+2}{n})\cdots (\frac{n+n}{n})=\frac{1}{n^n}(n+1)(n+2)\dots(n+n)$
please give some hint to proceed further.
 A: Consider the log of $[(1+\frac{1}{n})(1+\frac{2}{n})\dots(1+\frac{n}{n})]^{\frac{1}{n}}$:
$$\frac{1}{n}\sum_{k=1}^n \ln \left( 1+\frac{k}{n}\right).$$
The limit of this sum is equal to $\int_0^1\ln(1+x)dx$, so
$$
\begin{aligned}
\lim_{n\to\infty} [(1+\frac{1}{n})(1+\frac{2}{n})\dots(1+\frac{n}{n})]^{\frac{1}{n}}
&= \lim_{n\to\infty} \exp\left(\frac{1}{n}\sum_{k=1}^n \ln \left( 1+\frac{k}{n}\right) \right)\\
&= \exp\left(\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n \ln \left( 1+\frac{k}{n}\right) \right)\\
&= \exp \left( \int_0^1\ln(1+x)dx\right)
\end{aligned}$$
A: $$(\frac{n+1}{n})(\frac{n+2}{n})\dots (\frac{n+n}{n})=\frac{1}{n^n}(n+1)(n+2)\dots(n+n)$$
$$=\lim_{n\rightarrow \infty}\left[\frac{1}{n^n}\frac{(2n)!}{n!}\right]^{1/n}=\lim_{n\rightarrow \infty}\left[\frac{\sqrt{4\pi n}\cdot 2^{2n}\frac{n^{2n}}{e^{2n}}}{\sqrt{2\pi n}\cdot \frac{n^{2n}}{e^n}}\right]^{1/n}=\lim_{n\rightarrow \infty }[\sqrt{2}]^{1/n}\cdot \frac{4}{e}=\frac{4}{e}$$
Here I've used the Stirling approximation of $n!$,i.e- $n!=\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}$
