TITLE: Show that for any integer $n\ge 6$, the inequality $2^n>7n$ holds.
QUESTION [5 upvotes]: Step one: for case where $n=6$
$$7n <2^n$$
$$7(6)<2^6 \rightarrow 42<64$$.
Step two: Suppose for $n$ such that $7n<2^n$ is true. Now prove for $n+1$
$$(2)7n<(2)2^n$$
$$14n<2^{n+1}$$
But since $n<7n<2^n$, then $n+7<7(n+1)<2^n+7<2^{n+1}(?)$.
Then $7(n+1)<2^{n+1}$
I'm new in this induction process, so any help/tips for this problem would be really appreciated.

REPLY [4 votes]: $$2^n-7n=(1+1)^n-7n>1+n+\frac{n(n-1)}{2}+\frac{n(n-1)(n-2)}{6}-7n=$$
$$=\frac{n^3-37n+6}{6}=\frac{(n-6)(n^2+6n-1)}{6}\geq0.$$
Done!