What n is required to make n > a^n

[uart]

I'm looking for a bound on n ( that is, "n" as a function of "a") to satisfy,

$$n! \geq a^n$$

It doesn't have to be the lowest possible bound, preferably one that's relatively easy to prove.

Thanks.

 

[uart]

OK I've just found Sterling's formula, looking at that I can deduce that $$n = \lceil e\, a \rceil$$ will definitely work.

Does anyone have a bound that is very easy proof. Maybe something like n = 3a or even larger.

 

[Office_Shredder]

Well, up until n=a, you clearly can't get anywhere (and have a-1 more a's on the RHS than the LHS)

Then you add up an a (actually, a number greater than a against a)for every one added to n until you hit a²

Then you have a-2 a's on the RHS compared to the LHS. From then on, every time you add one to n, you multiply the LHS by something greater than a², and the RHS by a. So if n=a²+a (so you pick up a extra a's from the a² and larger terms to mitigate the a-1 you felll behind by at the start), then n! >= aⁿ

That should work. If it needs clarification, I'll try to write it out a bit neater

 

[uart]

Ok good, going as high as sqaures makes it pretty easy.

Take n = a^2 (for a>=2). You can just compare the terms (in the expansion of LHS and RHS) in pairs and show that every chosen pair-product on the LHS is greater than or equal to that of each RHS pair (which is of course always a*a on the RHS).

eg

1 * a^2 = a*a
2 * (a^2-1) > a*a
3 * (a^2-2) > a*a
...
k ^ (a^2+1-k) > a*a : { down to k = floor(a^2 / 2) }

That's the type of thing I was looking for, something that's like a "handwaving + by inspection" type of "proof". ;)