Finding $\lim_{x\to\infty} \frac {(x!)^{\frac 1 x}}{x}$ 
Find $\displaystyle \lim_{x\to\infty} \frac {(x!)^{\frac 1 x}}{x}$

I have no idea how to solve it, I can approximate it to be in $(0,1)$ by squeezing but getting to the solution $(\frac 1 e)$ seems like it would require a lot more. Is this an identity?  
Note: no integrals nor gamma function.
 A: While waiting for someone to find a duplicate a reminder of Stirling formula
$$x\to\infty\implies x!\sim \sqrt{2\pi x}\left(\frac{x}{e}\right)^x$$
So we have, when $x\to\infty$,
$$\frac{(x!)^{\frac{1}{x}}}{x}\sim \frac{1}{e}\left(\sqrt{2\pi}\right)^{\frac{1}{x}}x^{\frac{1}{2x}}\sim \frac{1}{e}$$
A: Just for variety, here's an alternative solution, based on the first simple observation in Terry Tao's notes on Stirling's formula.
We'll show that
$$ \frac{n^n}{n!} \le e^n \le 2(n+1)\frac{n^n}{n!} \tag{$\ast$} $$
Taking $n$th roots and squeezing yields the desired limit.  (The standard Riemann sum method — as, e.g., in the first stage in Tim Gowers' "Removing the magic from Stirling's formula" — gives the similar bounds $e\frac{n^n}{n!} \le e^n \le en\frac{n^n}{n!}$.)
Tao's observation gives us the lower bound in ($\ast$):
$$ \frac{n^n}{n!} \le \sum_{k=0}^\infty \frac{n^k}{k!} = e^n $$
Now let's consider how much we lost there by throwing away all terms of the series except the one with $k=n$.  First,
$$ \frac{n^k}{k!} \le \frac{n^{k+1}}{(k+1)!} \iff k\le n $$
and so the terms in the series $\sum_{k=0}^\infty \frac{n^k}{k!}$ increase from $k=0$ to $k=n$ and decrease thereafter.  In particular, for a longish head of the series we get
$$ \sum_{k=0}^{2n-1} \frac{n^k}{k!} \le 2n\cdot\frac{n^n}{n!} \tag{1} $$
In the remaining tail, we have $k\ge 2n$, and so
$$ \frac{n^{k+1}}{(k+1)!} = \frac{n}{k+1}\cdot\frac{n^k}{k!}
\le \frac12\cdot\frac{n^k}{k!} $$
so each term in the tail is less than half the preceding one.  Thus we can bound them by the geometric series $1+\frac12+\frac14+\dotsb$, getting
$$ \sum_{k=2n}^\infty \frac{n^k}{k!} \le 2\frac{n^{2n}}{(2n)!}
\le 2\frac{n^n}{n!} $$
which when combined with (1) gives the upper bound in ($\ast$).
A: Note that
$$
\frac{(n+1)!}{n!}=n+1
$$
and
$$
\frac{(n+1)^{n+1}}{n^n}=\left(1+\frac1n\right)^{n}(n+1)
$$
Therefore,
$$
\left.\frac{(n+1)!}{(n+1)^{n+1}}\middle/\frac{n!}{n^n}\right.=\left(1+\frac1n\right)^{-n}
$$
Since $\lim\limits_{n\to\infty}\left(1+\frac1n\right)^{-n}=e^{-1}$, Stolz-Cesaro applied to the logarithms says that
$$
\lim_{n\to\infty}\frac{(n!)^{1/n}}{n}=\lim_{n\to\infty}\left(\frac{n!}{n^n}\right)^{1/n}=e^{-1}
$$
A: Let us first consider the limit as $L$. Then, applying natural logarithm, we get,
$$\ln(L)=\lim_{x\to\infty}\frac{1}{x}\sum_{i=1}^x\left(\ln\left(\frac{i}{x}\right)\right)=\int\limits_0^1\ln(x)\,dx$$
Now, since the function $f(x)=\ln(x)$ is Riemann-integrable on $(0,1)$, we see, by the Riemann Sum argument that $L$ exists finitely. Next, we start to evaluate it.
$$L=\int\limits_0^1\ln (x)\,\mathrm dx=\left[x\ln(x)-x\right]_0^1\\ \implies \ln(L)=\left(\ln(1)-1\right)-\left(\lim_{x\to 0^+}(x\ln(x)-x)\right)=(-1)\\ \implies \ln(L)=(-1)\implies L=e^{-1}=\frac{1}{e}$$
A: Note this
$$  \left( \frac{x!}{x^x} \right)^{1/x} = (a_x)^{1/x}  $$
where $a_x = \frac{x!}{x^x}$ and then use the fact that 

$$ \lim_{x\to \infty} (a_x)^{1/x} = \lim_{x\to \infty} \frac{a_{x+1}}{a_x} $$

and the evaluation of limit will become easy
$$ \lim_{x\to \infty} \frac{a_{x+1}}{a_x} = \lim_{x\to \infty} \frac{1}{(1+1/x)^x} = \frac{1}{e}. $$
