A problem of limit: $\lim\limits_{n\to \infty}\left\{\frac{(n+1)^{n}}{n!} \right\}^{(1/n)}$ I need to find the limit of $$\lim\limits_{n\to \infty}\left\{\left(\frac{2}1\right) \left(\frac{3}2\right)^{2} \left(\frac{4}3\right)^{3} ... \left(\frac{n+1}n\right)^{n} \right\}^{(1/n)}$$

Cancelling out the same quantities in numerators and denominators, I have reached the step: $$\lim\limits_{n\to \infty}\left\{\frac{(n+1)^{n}}{n!} \right\}^{\frac1n}$$

Now I am stuck here. I have searched google for help, but found only the result for $\lim\limits_{n\to\infty}\left\{\frac{1}{n!} \right\}^{(\frac1n)}$. 

Which formulae/properties should I use now to proceed from this stage?

Edit: The options for the answer are: a) $e$, b) $1/e$, c) $\pi$, d) $1/\pi$
Taking logarithm on both the sides,
we get $$\log(y_n)=\log((\frac{(n+1)^n}{n!})^{\frac{1}{n}})$$
$$=\frac{1}{n}\{\log\frac{2}{1}+2\log\frac{3}{2}+...+n\log\frac{n+1}{n}\}.$$
Using Cauchy's First theorem on limits
$$\lim_{n\to\infty}\frac{1}{n}\{\log\frac{2}{1}+2\log\frac{3}{2}+...+n\log\frac{n+1}{n}\}=\lim_{n\to\infty}n\log\frac{n+1}{n}=1$$
Well, $e$ is the limit of $\left(\frac{(n+1)^n}{n!}\right)^{1/n}$ as $n$ goes to infinity. Indeed:

By Stirling's Approximation, $n!$ is $(1+o(1)\left(\frac{n}{e}\right)^n\sqrt{2\pi n}$. So $\left(\frac{(n+1)^n}{n!}\right)^{1/n}$ is

$[\sqrt{2\pi n}]^{1/n} \left(\left(\frac{n}{e(n+1)}\right)^n\right)^{1/n}$ which is $e^{-1}$.

Was that the question that was meant to be asked?