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.
 A: Set $f(n)=\frac{a^n}{n^a}$; we seek to prove $f(n)>1$ for all $n$ sufficiently large.  We rewrite this using one of the laws of exponents as
$$f(n)=\frac{e^{n \ln a}}{e^{a\ln n}}=e^{n\ln a-a\ln n}=e^{\ln n\left(\frac{n}{\ln n}\ln a - a\right)}$$
Hence we need only choose $n$ large enough that $\frac{n}{\ln n}\ln a - a>0$, i.e. such that $$\frac{n}{\ln n}>\frac{a}{\ln a}$$
It is easy to prove that $\frac{n}{\ln n}$ is an increasing function for $n\ge e$ (proof on request), hence for all $n>\max(a,e)$ this is true.

PS: The statement is false if $a\le 1$, so we must assume $a>1$.
A: Verify that the inequality fails if $a\le 1.$ So we assume $a>1.$
Let's first show $n < a^n$ for large $n.$ By the binomial theorem, for $n\ge 2,$
$$a^n = (1+(a-1))^n = 1 + n(a-1) + n(n-1)(a-1)^2/2 + \cdots \ge  n(n-1)(a-1)^2/2.$$
Clearly the last expression is $>n$ for large $n.$
To show $n^a < a^n$ for large $n,$ note that $a^{1/a}> 1.$ By the above, $n<(a^{1/a})^n$ for large $n.$ This is the same as saying $n<(a^n)^{1/a},$ or $n^a < a^n,$ for large $n.$
A: Take $\log$ from both sides $a \log n < n \log a $
Now lets check who  grows faster $\lim\limits_{n\to \infty}\frac{n \log a}{a \log n}=\infty$ and $\lim\limits_{n\to \infty}\frac{a \log n}{n \log a}=0$ 
A: It is enough to prove $\log(n^a)=a\log n<\log(a^n)n\log a$ for all sufficiently large $n$, i.e.
$$\frac{\log n}n<\frac{\log a}a,$$
which is true for all  sufficiently large $n$ since $\dfrac{\log n}n\to 0$ as $n\to\infty$.
A: 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}
$.
