Prove that $n^a < a^n$ for $a>1$ and $n$ big enough How can I solve this?
I'm trying to prove using limits but it's not working..
Thanks
 A: so start with for $a>1$ and $\forall n$ big enough we have
$$n^a < a^n \Leftrightarrow \log(n^a)<\log(a^n)\Leftrightarrow a\log(n)<n\log(a)\Leftrightarrow 1<\frac{\log(a)}{a}\frac{n}{\log(n)}
$$
and we fixed $a$, so $\frac{\log(a)}{a}=c>0$ is just a constant.
So in fact we have to show, that 
$$
1<c\frac{n}{\log(n)}
$$
holds for all constants $c>0$ and for all $n\ge n_0(c)$.
But this is ofcourse true, since 
$$
\lim_{n\rightarrow\infty}\frac{n}{\log(n)}=\infty 
\text{ if we consider }\bar{\mathbb{R}}:=\mathbb{R}\cup\{-\infty,+\infty\} 
$$ 
so it grows above each bound. 
bests
A: Let $x_{n} = a^{n}/n^{a}$ and then we have $$\frac{x_{n + 1}}{x_{n}} = \frac{a^{n + 1}}{(n + 1)^{a}}\cdot\frac{n^{a}}{a^{n}} = a\left(\frac{n}{n + 1}\right)^{a} \to a \text{ as } n \to \infty$$ Now $a > 1$ and hence we can choose a number $k$ with $1 < k < a$ and by the above limit there exists a positive integer $m$ such that $$\frac{x_{n + 1}}{x_{n}} > k$$ for all $n \geq m$. Thus we can see that $$\frac{x_{m + n}}{x_{m}} > k^{n} $$ for all positive integers $n$. Now we can see that $k > 1$ and hence $$k^{n} = (1 + (k - 1))^{n} \geq 1 + n(k - 1) $$ so that $k^{n} \to \infty$ as $n \to \infty$. It follows that $x_{m + n} > x_{m}k^{n}$ and hence $x_{m + n} \to \infty$ as $n \to \infty$. Thus $x_{n} \to \infty$ as $n \to \infty$.
It is now obvious that after a certain value of $n$ we will always have $x_{n} > 1$ and hence $a^{n} > n^{a}$ after a certain value of $n$. There is no need to go for logarithms and complicated limits related to them. Just simple limits coupled with definition of limit is sufficient to handle the problem.
A: My attempt at an inductive proof: 
Base Case: $n=1 \Rightarrow a^1 > 1$ Which is true by the initial condition
Assumption: $a^k > k^a$
Want to show: $a^{k+1} > (k+1)^a$
Multiply both sides of the assumption by $a$ to get: $a^{k+1} > ak^a$
Thus, it suffices to show that $ak^a > (k+1)^a$ 
Dividing both sides by the RHS gives: $a(\frac{k}{k+1})^a >1$
For $k$ large enough, the limit of $\frac{k}{k+1}$ is 1, so for sufficiently large $k$ the inequality reduces to $a>1$, which is true
I'm not sure if this is actually correct: can we consider a limit within an inductive proof? If not, is there another way we can finish the proof off? 
