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?

Yes, it is sort of obvious. Assume that $K\ge1$, then for any $\epsilon>0$, there is $N$ such that
$$(1+\epsilon)^n\ge 1+n\epsilon>K\text{ for }n\ge N.$$
What's less obvious is $n^{1/n}\to 1$.
