TITLE: Proof by induction that $2^n \gt n^k$
QUESTION [5 upvotes]: How do I prove using induction that if $k$ is a natural number, then $2^n \gt n^k$ for all $n \geq k^2 + 1$, where $n$ is also a natural number?

REPLY [5 votes]: If you really want to use induction, then you might use this lemma I came up across with by playing a little with your inequality :)
Lemma 1 :
$ \left( 1+\frac{1}{n} \right)^k < 2 $ for $ k \geq 1 $ ( that means $n \geq 2$ )
Proof of Lemma 1 :
By the convexity of $exp$, we know that $1+\frac{1}{n} < e^{\frac{1}{n}}$, and so $ \left( 1+\frac{1}{n} \right)^k < e^{\frac{k}{n}} $.
It suffices then to prove that $ \ln 2 - \frac{k}{n} > 0 $, which is immediate because $\ln 2 - \frac{k}{n} \geq \ln 2 - \frac{\sqrt{n-1}}{n} = \ln2 - \sqrt{\frac{1}{n} - \frac{1}{n^2}}$ and $x \to \frac{1}{x}-\frac{1}{x^2}$ is a decreasing function on $[2,+\infty[$, so $\ln 2 - \sqrt{ \frac{1}{n} - \frac{1}{n^2} } \geq \ln 2 - \frac{1}{2} > 0$, which completes the proof.