Inspired by Harsh Shrivastava

We apply a variation of Stolz–Cesàro theorem:

If the sequence $(x_n /x_{n-1} )_{n=1}^\infty$ has a limit, then $\displaystyle \lim_{n\to\infty} \sqrt[n]{x_n} = \lim_{n\to\infty} \frac{x_n}{x_{n-1}}$ .

Because $\frac {n+1}{(n!)^{1/n}} = \sqrt[n]{ \frac{(n+1)^n}{n!}}$ , let $x_n = \frac{(n+1)^n}{n!}$ , then $\displaystyle \lim_{n\to\infty}\frac{x_n}{x_{n-1}} = \lim_{n\to\infty}\left(1 + \frac1n\right)^n = e$ , which is finite.

Thus Stolz–Cesàro theorem is applicable. With the limit equals to $e$ .

Split it like this:

$\displaystyle \lim_{n \to \infty} \frac{n}{(n!)^{1/n}} + \frac{1}{(n!)^{1/n}}$

Let $y$ = $\displaystyle \lim_{n \to \infty} (\frac{n^n}{n!})^{1/n} + 0$

Taking log both the sides $\displaystyle\log y = \lim_{n \to \infty} \frac{1}{n}\log(\frac{n}{n-0}. \frac{n}{n-1}. \frac{n}{n-2}......\frac{n}{n-(n-1)})$

$\displaystyle \log y = \lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{n-1} \log(\frac{1}{1-\frac{r}{n}})$

$\displaystyle \log y = \int\limits_0^1 \log(\frac{1}{1-x}) dx$

$\log y = 1$ $y = \boxed{e}$

We can use Stirling's approximation in the form

$\ln(n!) = n\ln(n) - n + O(\ln(n)).$

With $L$ being the given limit, we then have that

$\ln(L) = \lim_{n \rightarrow \infty} (\ln(n + 1) - \frac{1}{n}*\ln(n!)) =$

$\lim_{n \rightarrow \infty} (\ln(n + 1) - \ln(n) + 1 - \frac{1}{n}*O(\ln(n))) =$

$1 + \lim_{n \rightarrow \infty} (\ln(1 + \frac{1}{n})) - \lim_{n \rightarrow \infty} (\frac{1}{n}*O(\ln(n))) = 1 + \ln(1) - 0 = 1.$

(Note that, by L'Hopital's rule,

$\lim_{x \rightarrow \infty} \frac{\ln(x)}{x} = \lim_{x \rightarrow \infty} \dfrac{\frac{1}{x}}{1} = 0,$

the result then holding true for the analogous discrete limit.)

This implies that $L$ itself is $e^{1} = \boxed{e}.$