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) $1/\pi$
 A: 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
$$\lim_{n\to\infty}\frac{1}{n}\{\log\frac{2}{1}+2\log\frac{3}{2}+...+n\log\frac{n+1}{n}\}=\lim_{n\to\infty}n\log\frac{n+1}{n}=1$$
A: As you say, $n \to 0$ is rather unusual. So, let us consider the case of $x \to 0$ for
$$A=\left(\frac{(x+1)^x}{x!}\right)^{\frac{1}{x}}=\frac{x+1}{ \Gamma (x+1)^{\frac{1}{x}}}\implies \log(A)=\log(x+1)-\frac 1x\log (\Gamma (x+1))$$ Now, close to $x=0$ (have a look here),
$$\log (\Gamma (x+1))=-\gamma  x+\frac{\pi ^2 x^2}{12}+O\left(x^3\right)$$
$$\frac 1x\log (\Gamma (x+1))=-\gamma  +\frac{\pi ^2 x}{12}+O\left(x^2\right)$$ making $$\log(A)=\gamma +\left(1-\frac{\pi ^2}{12}\right) x+O\left(x^2\right)$$ Continuing with Taylor
$$A=e^{\log(A)}=e^\gamma \left(1+\left(1-\frac{\pi ^2}{12}\right) x+O\left(x^2\right) \right)$$
For illustration purposes, let $x=10^{-k}$ and compare the values
$$\left(
\begin{array}{ccc}
k & \text{approximation}& \text{exact} \\
 0 & 2.097271488 & 2.000000000 \\
 1 & 1.812692325 & 1.811281476 \\
 2 & 1.784234409 & 1.784219608 \\
 3 & 1.781388617 & 1.781388468 \\
 4 & 1.781104038 & 1.781104036 \\
 5 & 1.781075580 & 1.781075580
\end{array}
\right)$$
A: Take the $\log$ and you find $$\log(n+1) - \log(n) + \log(n) - \frac{1}{n}\sum_{k=1}^n \log(k) = \log(1+\frac{1}{n}) - \frac{1}{n}\sum_{k=1}^n \log(\frac{k}{n}) \rightarrow -\int_0^1 \log(x) \mathrm d x = 1$$
as $$\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$
A: 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?
