Everything boils down to the n times use of the following elementary limit, namely: $\lim_{x\to0}\frac{e^{x}-1}{x}=1$
Consequently, by expanding the sum we have that:
$\lim_{n\to\infty}\frac{e^\frac{1}{n+1}-1}{\frac{1}{n+1}} \frac{1}{n+1} + \lim_{n\to\infty}\frac{e^\frac{1}{n+2}-1}{\frac{1}{n+2}} \frac{1}{n+2}+ \cdots = \lim_{n\to\infty}\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} = $ $\lim_{n\to\infty}{\gamma}+\ln{2n}-{\gamma}-\ln{n}= \ln{2}.$ Also notice that i've just applied another well-known limit: $\lim_{n\to\infty} 1+\frac1{2}+\cdots+\frac{1}{n}-\ln{n}={\gamma}$ $\tag{$\gamma$ is Euler-Mascheroni constant}$
The proof is complete.