<s>[INST] $2^n > n^4$ proof by induction This is what I came up with so far:
Inductive step: assume $2^n > n^4$.
Need to prove $2^{n+1} > (n+1)^4$
$$
2^{n+1} = 2 \cdot 2^n > 2 \cdot n^4\\
(2 \cdot n^4)^{1/4} = (2)^{1/4} \cdot n > n+1 \implies 2n^4 > (n+1)^4 \implies 2^n > (n+1)^4
$$
Is there a better way to solve this problem?
 [/INST] Here is a nice little non-induction proof that $n^4\lt2^n$ for $n\gt16$.
Note that
$$n^4=\left(\sqrt n\over2\right)^8\cdot2^8\cdot1^{n-16}$$
So if $n\gt16$, the arithmetic-geometric mean inequality tells us
$$\sqrt[n]{n^4}\lt{8\cdot\left(\sqrt n\over2\right)+8\cdot2+(n-16)\cdot1\over n}={4\over\sqrt n}+1\lt2$$
hence $n^4\lt2^n$.  
(Remark: for $n=16$ both inequalities in the displayed expression become equalities.  In particular, all $16$ terms in the AGM product $\left(\sqrt4\over2\right)^8\cdot2^8$ are equal to $2$.  For $n\lt16$, the AGM connection falls apart, as it must, since the inequality goes the other way.)
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