Hi, what can be the limiting values of n for which $\displaystyle n!>a^n$ for any given a?
Thanks.
I haven't worked this out completely, but I suspect you might find Stirling's approximation useful:
$\displaystyle \displaystyle n! \approx \sqrt{2 \pi n} \left ( \frac{n}{e} \right ) ^n$
That would give your problem the following form:
$\displaystyle \displaystyle \sqrt{2 \pi} n^{n + 1/2} > (ea)^n$
-Dan
Well first thing first, what I wrote is the lowest n possible (there isn't a maximum cause you can always pick another one greater by adding 1 to the preceding number).
I am unaware of a closed analytical formula for what I wrote, I guess only Stirling approximation can work here (if you want to find some closed formula though still an approximation).