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Is it possible to calculate the maximum interface value for $n$ so that $n!<100^{n-1}$, without using computer or calculator? I thought of using Sterling but

$$\ln n! = n \ln n - n + 1/2 \ln (2 \pi n) + 1/(12n) - ... $$ Needs calculating $\ln n$, not so easy seems

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    Yes, it is possible: using a lot of ink and paper. – 2017-01-10
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    Steirling formula is good to replace instead of $n!$ – 2017-01-10
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    @MyGlasses pls see my post updated – 2017-01-10
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    In this case not using a calculator seems rather senseless. Why are you trying to compute this by hand? – 2017-01-10

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I'd take $\ln$ of both sides. Then the left side is

$$\sum_1^n \ln n \approx \int_1^n \ln x \; dx = n\ln n - n +1$$

which needs to be less than $(n-1)\ln 100$. So

$$\frac{n}{n-1}\ln n - 1 < \ln 100 = 4.605.$$

You have to grind out $\ln 100$ using Taylor series. Then use $\frac{n}{n-1} \approx 1$ and we have $n\approx e^{5.605}.$ Another Taylor series calculation gives $n = 271.78.$ The right answer is $n=270$, so we got pretty dang close.