Geometric Mean of "All" Numbers between $1$ and $2$

$\displaystyle\lim_{N\to\infty}{\left(\displaystyle\prod_{a=0}^{N}(1+\frac{a}{N})\right)}^{\frac{1}{N+1}}$ $=\exp{\ln{\displaystyle\lim_{N\to\infty}{\left(\displaystyle\prod_{a=0}^{N}(1+\frac{a}{N})\right)}^{\frac{1}{N+1}}}}$ where $\exp{x}=e^x$ $=\exp{\displaystyle\lim_{N\to\infty}\ln{\left(\displaystyle\prod_{a=0}^{N}(1+\frac{a}{N})\right)}^{\frac{1}{N+1}}}$ $=\exp{\displaystyle\lim_{N\to\infty}\frac{1}{N+1}\ln{\left(\displaystyle\prod_{a=0}^{N}(1+\frac{a}{N})\right)}}$

Using the rules of logarithms, $=\exp{\displaystyle\lim_{N\to\infty}\frac{1}{N+1}\displaystyle\sum_{a=0}^{N}\ln{(1+\frac{a}{N})}}$

Switch to integral notation with $x=\frac{a}{N}$ and $dx=\frac{1}{N}$ : $=\exp{\left(\displaystyle\int_{0}^{1}\ln{(1+x)}\ dx\right)}$ $=e^{[(1+x) \ln{(1+x)}-1-x]_{0}^{1}}$ $=e^{2\ln(2)-1}$ $=\frac{2^2}{e}$ $=\frac{4}{e}$