Proving that the limit $ \lim\limits_{n\rightarrow \infty} (n!)^{\frac{1}{n}}$ diverges to infinity I came across this limit in some context:
$$ \lim\limits_{n\rightarrow \infty} (n!)^{\frac{1}{n}}$$
I could only say that $n! > n$ implies the limit is greater than or equal to $1$. However, the result seems to be infinity. I do not know how to arrive at this result.
Any ideas?
Based on the answer and the comment below, I wonder if it is possible to prove this using elementary Calculus?
 A: The series $\displaystyle \sum_{n= 0}^\infty\frac{x^n}{n!}$ converges for all $x$, so $\displaystyle \lim_{n\to\infty}\frac{x^n}{n!}=0$ for any $x$. This means in particular that, given any fixed $M$, we have $n!>M^n$ for $n$ large enough. Thus $(n!)^{1/n}>M$ for $n$ large enough. Since $M$ was arbitrary, $\lim_{n\to\infty}(n!)^{1/n}=\infty$.
A: You can do it like this. Fix an arbitrary $n_0$ and see what happens when $n > n_0$.
Clearly,
$$
n! = n_0! \cdot (n_0+1) (n_0+2) \ldots (n).
$$
It follows that
$$
n! \geq n_0! \cdot n_0^{n - n_0},
$$
and
$$
\sqrt[n]{n!} \geq \sqrt[n]{n_0! \cdot n_0^{n-n_0}} = \sqrt[n]{\frac{n_0!}{n_0^{n_0}}} \cdot n_0.
$$
Now let's see what happens when $n \to \infty$. The right hand side converges to $n_0$. So, when $n$ is large enough, we can say, for example, that $\sqrt[n]{n!} \geq \frac{1}{2}n_0$.
But, since $n_0$ is arbitrary, we see that for any positive $C$ we will have $\sqrt[n]{n!} \geq C$ for all large enough $n$. It means that $\sqrt[n]{n!}$ tends to infinity.
A: The most elementary proof I know goes like this: note that $n! \ge n(n-1)\cdots \frac n2 \ge (\frac n2)^{n/2}$. (The middle step needs fixing if $n$ is odd, but the inequality $n! \ge (\frac n2)^{n/2}$ still stands.) Therefore
$(n!)^{1/n} \ge (\frac n2)^{1/2}$, which tends to infinity; hence $(n!)^{1/n}$ itself tends to infinity.
A: use  this well kown
$$\left(\dfrac{n}{e}\right)^n<n!<e\left(\dfrac{n}{2}\right)^n$$
or you can use
$$n!\approx\left(\dfrac{n}{e}\right)^n\sqrt{2n\pi}$$
