$\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{Lt}}{\left[\left(1+\frac{1}{\mathrm{n}}\right)\left(1+\frac{2}{\mathrm{n}}\right)\dots \left(1+\frac{\mathrm{n}}{\mathrm{n}}\right)\right]}^{1/\mathrm{n}}=$

$$\begin{gathered} \underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{Lt}}{\left[\left(1+\frac{1}{\mathrm{n}}\right)\left(1+\frac{2}{\mathrm{n}}\right)\dots \left(1+\frac{\mathrm{n}}{\mathrm{n}}\right)\right]}^{1/\mathrm{n}} \\ \mathrm{s}=\underset{\mathrm{n}\to \infty }{\lim }{\left[\right(1+\frac{1}{\mathrm{n}})+(1+\frac{2}{\mathrm{n}})+.....+(1+\frac{\mathrm{n}}{\mathrm{n}}\left)\right]}^{\frac{1}{\mathrm{n}}} \end{gathered}$$
$\mathrm{logs} = \underset{\mathrm{n}\to \infty }{\lim }\frac{1}{\mathrm{n}}\left[\log \right(1+\frac{1}{\mathrm{n}})+\log (1+\frac{2}{\mathrm{n}})+...+\log (1+\frac{\mathrm{n}}{\mathrm{n}}\left)\right]$
$\mathrm{logs} = \underset{\mathrm{n}\to \infty }{\lim }\frac{1}{\mathrm{n}}\underset{ }{\overset{ }{\sum _{\mathrm{r}=1}^{\mathrm{n}}\left[\log \right(1+\frac{\mathrm{r}}{\mathrm{n}}\left)\right]}}$
$\mathrm{logs} = {\int }_0^1\log (1+\mathrm{x})\mathrm{dx}$
$$\begin{gathered} \mathrm{logs} = \log (1+\mathrm{x}).\mathrm{x}{]}_0^1-{\int }_0^1\frac{1}{1+\mathrm{x}}.\mathrm{xdx} \\ \mathrm{logs} = \log 2-{[\mathrm{x}-\log (1+\mathrm{x}\left)\right]}_0^1 \\ \mathrm{logs} = \log 2-[1-\log 2] \\ \mathrm{logs} = \log 2-1+\log 2 \Rightarrow \mathrm{s} =\mathrm{u}-{\mathrm{e}}^{-1} \Rightarrow \mathrm{s} = \frac{\mathrm{u}}{\mathrm{e}} \end{gathered}$$