Computing the $\lim_{n\to\infty} n^{\frac{1}{n!}}$? I want to find the following limit
$$\lim\limits_{n\rightarrow \infty} n^{\frac{1}{n!}}$$ 
I tried solution as follows:
Let $L=n^{\frac{1}{n!}}$ this implies $\log L=\frac{\log n}{n!}$ which is $\frac{\infty}{\infty}$ form as $n\rightarrow \infty.$ Don't know how to proceed. Help required
 A: Hint:
$$1 \le n^{1/n!} \le (n!)^{1/n!}$$
A: HINT:
$$\lim_{n\to\infty}n^{\frac{1}{n!}}=\lim_{n\to\infty}\exp\left(\ln\left(n^{\frac{1}{n!}}\right)\right)=\lim_{n\to\infty}\exp\left(\frac{1}{n!}\ln\left(n\right)\right)=\exp\left(\lim_{n\to\infty}\frac{\ln(n)}{n!}\right)$$
A: I am writing solution on the hint provided by @Crostul
We have $0<\frac{\log n}{n!}<\frac{1}{(n-1)!}.$  Therefore $0\leq \lim\limits_{\lim\rightarrow\infty} \frac{\log n}{n!}\leq 0$ that is $L=1.$
A: Try with
$$n^{\frac{1}{n!}} = e^{\frac{1}{n!}\ln(n)}$$
Then your limit will be in the form $\frac{\infty}{\infty}$. The $\ln(n)$ function grow reeeeeeeeeeally slow with respect to $n!$ thus the fraction is zero and you get $e^0 = 1$.
Limiti is $1$.
A: $n^{\frac{1}{n!}} = e^{\frac{1}{n!} \ln (n)},$
$ \lim_{x \to \infty} \frac{1}{ {n!}} \ln(n)\leq \lim_{x \to \infty} \frac{1}{n} \ln(n) =\lim_{x \to \infty} \frac{\ln(n)}{n}=\lim_{x \to \infty} \frac{\frac{1}{n}}{1} = 0.  $
So  $n^{\frac{1}{n!}} = e^{\frac{1}{n!} \ln (n)}=e^0 = 1$ as $n \to \infty.$
A: Just to give a non-log-invoking answer, note that for $n\gt1$ we have
$$n^{n+1}-n=n(n^n-1)\gt1$$
and thus $n^{n+1}\ge n+1\ge1$ from which it follows that
$$1\le(n+1)^{n!}\le(n^{n+1})^{n!}=n^{(n+1)!}$$
and thus
$$1\le(n+1)^{1/(n+1)!}\le n^{1/n!}$$
so the sequence $n^{1/n!}$ (for $n\gt1$) is a decreasing sequence of real numbers bounded below by $1$.  Thus there is a limit $L\ge1$.  We need now only show that $L=1$.
One way to do this is to note that for $n\ge2$ we have
$$(2n)^{2/(2n)!}\le(n^2)^{2/(2n)!}=n^{4/(2n)!}=(n^{1/n!})^{4/((2n)(2n-1)\cdots(n+1))}\le(n^{1/n!})^{2/n}\le n^{1/n!}$$
(where $n\ge2$ gets used in the first and last inequalities). Thus
$$1\le L\le L^2\le((2n)^{1/(2n)!})^2\le n^{1/n!}\to L$$
It follows that $L^2=L$, hence $L=1$ (since $L\ge1$ implies $L\not=0$).
Remark: One can give a similar first-principles proof that $n^{1/n}\to1$ as $n\to\infty$, but the inequality $(n+1)^{1/(n+1)}\le n^{1/n}$ is a little trickier to establish. (For one thing, it doesn't kick in until $n\gt2$, since $\sqrt[3]3\gt\sqrt2$.) The factorial here makes life easier.
