What is the smallest integer $n$ such that $n! > 100^n$?

Question

I know the answer is more than $100$, because $100!$ is clearly less than $100^{100}$. And I know the answer is less than $10000$, because ($10000 \cdot 9999 \cdot 9998 \cdot \dots\cdot 9901) \cdot (100 \cdot 99 \cdot 98 \cdot \dots \cdot 1)$ is greater than $100^{200}$. But I don't know how to find the answer. Can this be done without testing every value?

Answer 1

Stirling says $n! \sim \sqrt{2\pi} n^{n+1/2} e^{-n}$. $(n/e)^n = 100^n$ for $n = 100 e \approx 271$. Not far off from the actual value, but getting something much better without technology might be difficult.

EDIT: I take that last sentence back slightly. Generalizing to $n! = t^n$, a better solution asymptotically is $n \sim t e - \ln(2\pi t)/2 - 1/2$, as this makes $ \ln(n!) - \ln(t^n) = O(\ln(t)^2/t)$.

For $t = 100$ this would be $n = 100\; e - \ln(200 \pi)/2 - 1/2 \approx 268.107$. Indeed this is quite close to the actual solution of $\Gamma(x+1) = 100^x$, which is approximately $268.087$, close enough so the next higher integer, $269$, is the correct value.

Of course some low-tech "technology" would be convenient for numerically evaluating $100 e - \ln(200 \pi)/2 - 1/2$.

I hope Rob Arthan likes this version better.