PhD admission product $\lim_{n\to 0}\left(\frac21\left(\frac32\right)^2\left(\frac43\right)^3\cdots\left(\frac{n+1}{n}\right)^n\right)^{1/n}$ Hello there I saw this problem (#3) here:
http://www.sau.int/admission/2018/samplepapers/PAM.pdf

$$L=\lim_{n\to 0} \left( \frac{2}{1}\left(\frac{3}{2}\right)^2\left(\frac{4}{3}\right)^3...\left(\frac{n+1}{n}\right)^n\right)^\frac{1}{n}$$
  The choices for the answer are $e$, $\pi$, $\frac{1}{e}$, $\frac{1}{\pi}$.

If we take the logartihm on both sides we get: $$\ln L=\frac{1}{n}\sum_{k=1}^{n} k\ln\frac{k+1}{k}$$ thus by telescoping $$\ln L= \ln\left(\frac{n+1}{(n!)^{\frac{1}{n}}}\right)$$ and now using wolfram I get the answer to be $L=e^\gamma$, which is not one of the choices. 
Where did I go wrong? And could you share the correct solution? 

$$\dfrac{n+1}{(n!)^{\frac{1}{n}}} \approx \dfrac{n+1}{\sqrt{2 \pi n}^{1/n}\dfrac n e} \to e$$
