I was reading a little from a book on Real Analysis (which I'm not that far into yet), and I came across a problem: prove $2^n \ge n^2, \forall n >4\in \Bbb N$. Though "challenging", I was able to prove this problem with induction without much struggle. I then considered proving this same problem for any integer '$a$' in replace of $2$. I.e. prove:
$$a^n \ge n^a$$
where $a,n \in \Bbb N$. The first thing I decided to do was look for the smallest '$n$' that would actually work for a given '$a$'. I did this by simply solving for $n$:
$$\begin {align} a^n \ge& n^a\\ n^{1 \over n} \ge& a^{1 \over a}\\ {1 \over n}\ln n \ge& \ln a^{1 \over a}\\ \ln n e^{-\ln n} \ge& \ln{a^ {1 \over a}}\\ -\ln n e^{-\ln n} \ge& -\ln{a^ {1 \over a}}\\ -\ln n \ge& W_0(-\ln{a^{1 \over a}})\\ n \ge& e^{-W_0(-\ln a^{1 \over a})} \end {align}$$
where $a \in \Bbb N$ and $W_0(x)$ is Lambert's $W$ Function.
My first question is: "Is this a valid constraint to put on $n$?", and my second and third questions are: "Is this a valid method for proving the inequality for values of $n$ larger than or equal to what I just derived? And if it isn't, how do I prove it using induction?"
Not by induction, but one way to go about it. Note that $$ \left(\frac{t}{\ln (t)} \right )' = \frac{(-1 + \ln(t))}{\ln^2(t)}$$ Thus $t /\ln t$ is strictly increasing for $t > e$. Therefore, for $a,t > e$ and $t \ge a$ $$ \frac{t}{\ln t} \ge \frac{a}{\ln a}$$ $$\Rightarrow a^{t} \ge t^{a} $$ and for $a,t >e$ and $t < a$, we have $$ \frac{t}{\ln t} < \frac{a}{\ln a}$$ $$\Rightarrow a^{t} < t^{a} $$ Thus for intgers $a,n \ge 3$, if $n\ge a,$ we have $a^n \ge n^a$ and if $n < a$, we have $a^n <n^a$.