Proof problem: show that $n^a < a^n$ for all sufficiently large n I would like to show that $n^a < a^n$ for all sufficiently large $n$, where $a$ is a finite constant.  
This is clearly true by intuition/graphing, but I am looking for a rigorous proof.  Can anyone help me out? Thanks.

What you want is
$n^{1/n} < a^{1/a}$.
Since $x^{1/x}$
has a max at $x=e$
and is decreasing for
$x > e$,
$n^{1/n} < a^{1/a}$
if $e \le a < n$.
If
$0 < a < 1$,
then,
since
$a^{1/a} < 1$
and
$n^{1/n} > 1$ for $n > 1$,
$n^{1/n} > a^{1/a}$
so what you want does not happen.
If $a > 1$,
since
$a^{1/a} > 1$
and
$n^{1/n} \to 1$
as
$n \to \infty$,
what you want
does happen for large enough $n$.
To get a simple bound on
$n^{1/n}$:
By Bernoulli's inequality,
$(1+n^{-1/2})^n
> 1+n(n^{-1/2})
=1+n^{1/2}
\gt n^{1/2}
$.
Raising both sides
to the $2/n$ power,
$(1+n^{-1/2})^2
\gt n^{1/n}
$
or
$n^{1/n}
< (1+n^{-1/2})^2
< 1+3n^{-1/2}
$
for $n > 1$.
Therefore
a sufficient condition
for
$n^{1/n} < a^{1/a}$
is
$a^{1/a}
> 1+3n^{-1/2}
$
or
$3n^{-1/2}
< a^{1/a}-1
$
or
$9/n
< (a^{1/a}-1)^2
$
or
$n
> \dfrac{9}{(1+a^{1/a})^2}
$.
