If we have the stirlings formula: $$N!=\sqrt{2\pi N}(\frac {N}{e})^N$$
And I am asked to find how big the N must be so our error is less then 1 %.
Because the error is in percentage, I am considering the relative error, which is the ration between the absolute and the actual value :
$$\frac{|N!-\sqrt{2\pi N}(\frac {N}{e})^N |}{|N!|}=0.01$$
But I don't know how to proceed in finding the N value for this case of an error (which should be around 9, which is the answer to this)
Well, using the formal definition we must solve:
$$1\space\text{%}\le100\space\text{%}\cdot\left|\frac{\text{n}!-\sqrt{2\pi\text{n}}\cdot\left(\frac{\text{n}}{e}\right)^\text{n}}{\text{n}!}\right|\tag1$$
And we find that:
$$\text{n}\ge9\tag2$$