Stirling formula, O-Notation In lecture we proof that
:  $ \sqrt{2 \pi n} (\frac{n}{e})^n< n! \le \sqrt{2 \pi n} (\frac{n}{e})^n e^{\frac{1}{12n}}$
But how we came form this to the formula $n!= \sqrt{2 \pi n} (\frac{n}{e})^n (1+ O(1/n)) $ with $0< O(1/n) \le \frac{1}{12n}$
I can rewirte the second formula as that  $\lim\sup_{n\rightarrow \infty}| \frac{\frac{n!}{\sqrt{2 \pi n}*(\frac{n}{e})^n} -1}{1/n}|$ exists, what means that $\frac{\frac{n!}{\sqrt{2 \pi n}*(\frac{n}{e})^n} -1}{1/n}$ is bounded. 
When i use the first formula i get $0=\frac{1-1}{n} \le \frac{\frac{n!}{\sqrt{2 \pi n}*(\frac{n}{e})^n} -1}{1/n} \le \frac{1}{12}*\frac{(e^{\frac{1}{12n}}-1)}{1/12 n}$
I know $\lim_{n\rightarrow \infty} \frac{1}{12}*\frac{(e^{\frac{1}{12n}}-1)}{1/12 n}= \frac{1}{12}*1=\frac{1}{12}$ so the right side is bounded and so the limessuperior exists.
I need to show that $ \frac{n!}{\sqrt{2 \pi n} (\frac{n}{e})^n} -1 \le \frac{1}{12n}$, that means $ \frac{\frac{n!}{\sqrt{2 \pi n} (\frac{n}{e})^n} -1}{\frac{1}{n}} \le \frac{1}{12}$. But this formula i only get for $n$ big enough and not for all n?

You have proved that, as $n\to \infty$,
$$
\sqrt{2 \pi n} \:\left(\frac{n}{e}\right)^n< n! \le \sqrt{2 \pi n} \:\left(\frac{n}{e}\right)^ne^{\large \frac{1}{12n}}\tag1
$$ from wich one deduces that, as $n \to \infty$,
$$
1< \frac{n!}{\sqrt{2 \pi n} \:\left(\frac{n}{e}\right)^n} \le e^{\large \frac{1}{12n}} \tag2
$$ by the Taylor series expansion, as $n \to \infty$, one has
$$
e^{\large \frac{1}{12n}}=1+O\left(\frac1n\right)\tag3
$$ giving
$$
1< \frac{n!}{\sqrt{2 \pi n} \:\left(\frac{n}{e}\right)^n} \le 1+O\left(\frac1n\right) \tag4
$$ or, as $n \to \infty$,

$$
n!= \sqrt{2 \pi n} \:\left(\frac{n}{e}\right)^n\left(1+ O\left(\frac1n\right)\right) \tag5
$$ 

as announced.
