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
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.