How can you prove that n! majorizes a^n (a is some real constant) by taking the limit of (a^n)/n! as n tends to infinity *without* introducing the gamma function?

Question

anonymous · Answer

ok i give up
what does "majorize" mean?  the limits as $n\to \infty$ of $\frac{a^n}{n!}=0$ ?

anonymous · Answer

Essentially show a^n << n! by taking that limit

anonymous · Answer

this is the series for $e^a$ which converges for all $a$
since it is convergent, the terms must go to zero
not much a proof i guess, but certainly true

anonymous · Answer

actually i like that proof, unless you do not get to use that fact
you could say that after $n>a$ each fraction is less than one except for the finite ones at the beginnig

anonymous · Answer

I'm not entirely sure I would get away with that. Thanks for your help anyway though, I think I may have it

anonymous · Answer

the second part should work though right?  after $n>a$ you have 
$$\frac{a}{1}\times \frac{a}{2}...\times \frac{a}{n}$$ and $\frac{a}{n}<1$ so after that, each fraction is less than one, and there are an infinite number of them 
only the first ones with $a\geq n$ are larger than one

anonymous · Answer

Yes, so I've named the product of the first a terms M.
So 0 < a^n/n! < M(a/n)
and M(a/n) -> 0 as n → ∞
therefore a^n/n! → 0 by the squeeze (sandwich) theorem