sequence: use of stirling formula I want to use the Sterling formula which says that:
$lim_{n \rightarrow \infty}\dfrac{n!}{\sqrt{2*\pi}n^{n+1/2}*e^{-n}}=1$
I want to use it to show that $\lim_{n \rightarrow \infty}(n!)^{1/n}=\infty$. However, there is one thing I have to do, which I do not have the justification for, can it be justified?, here it is:
$lim_{n \rightarrow \infty}(n!)^{1/n}=lim_{n \rightarrow \infty}(n!)^{1/n}\dfrac{(\sqrt{2*\pi}n^{n+1/2}*e^{-n})^{1/n}}{(\sqrt{2*\pi}n^{n+1/2}*e^{-n})^{1/n}}$
$=lim_{n \rightarrow \infty}\left(\dfrac{n!}{\sqrt{2*\pi}n^{n+1/2}*e^{-n}}\right)^{1/n}*(\sqrt{2*\pi}n^{n+1/2}*e^{-n})^{1/n}$.
Now comes my problem.  It is easy to see that this expression goes to infinity if we know that:
$lim_{n \rightarrow \infty}\left(\dfrac{n!}{\sqrt{2*\pi}n^{n+1/2}*e^{-n}}\right)^{1/n}=1$. But do we nececarrily know that even though what is inside that parenthesis goes to 1? 
I guess a similar question is that if $a_n\rightarrow K, K>0$, then $(a_n)^{1/n}\rightarrow 1$?, but is this really obvious?, if so, can you please show me a delta-epsilon proof of this?

I would rather make an estimate, and worry about limits later.  
Assuming the Stirling formula, we find that if $n$ is large enough, then 
$$n!\gt \frac{1}{2}\sqrt{2\pi}n^{n+1/2} e^{-n}\gt n^ne^{-n}.$$
Taking $n$-th roots, we find that if $n$ is large enough then
$$(n!)^{1/n}\gt \frac{n}{e}.$$
This settles the question. 
Remark: The Stirling formula is overkill. Note that 
$$(n!)^2 =[(n)(1)][(n-1)(2)][(n-3)(3)]\cdots [(1)(n)].$$
The products $(n-k)(k)$ are all $\ge n$. It follows that $n!\ge n^{n/2}$, and therefore $(n!)^{1/n}\ge n^{1/2}$. 
