Show that for any integer $n\ge 6$, the inequality $2^n>7n$ holds. 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.
 A: We will prove the assertion by induction on $n$. 

The assertion is true for $n=6$. $\checkmark$
Suppose that the assertion is true for $k=n$; $\color{Blue}{7n<2^n}$.
on the otherhand we know that $\color{Red}{7}<42\leq7n<\color{Red}{2^n}$; so we have :
$$ 
\color{Green}{7(n+1)}= 
\color{Blue}{7n} 
+ 
\color{Red}{7} 
< 
\color{Blue}{2^n} 
+ 
\color{Red}{2^n} 
= 
\color{Green}{2^{n+1}} ,$$ 
so the assertion is true for $\color{Green}{k=n+1}$.
A: You made a very good begin. 
I only don't see why you need $n+7 < 7(n+1)$. 
You just need $7(n+1) < 2^n+7 < 2^{n+1}$ and you are done.
You marked $2^n+7 < 2^{n+1}$ with a question mark. It is certainly something you should prove. In general, when you don't immediately see why something is true, you have to prove it. You can do this by subtracting $2^n$ from both sides. Since you ask for a hint, I hid the complete answer below.

 Since $n\geq 6$, we have $2^n \geq 64 > 7$, so $2^n+7 < 2^n + 2^n = 2^{n+1}$.

A: $$2^n-7n=(1+1)^n-7n>1+n+\frac{n(n-1)}{2}+\frac{n(n-1)(n-2)}{6}-7n=$$
$$=\frac{n^3-37n+6}{6}=\frac{(n-6)(n^2+6n-1)}{6}\geq0.$$
Done!
A: you have to show:
$$2^{n+1}>7(n+1)$$ multiplying $$2^n>7n$$ by $2$ we get
$$2^{n+1}>14n$$ and it is $$14n>7n+7$$ and this is true since $$n\geq 6$$
A: $2^n=(1+1)^n=2+2n+n(n-1)+f(n)>n(n+1)$, since $f(n)>0$, for $n\geq 3$
Since $n\geq 6$, we have $2^n>n(n+1)\geq 7n\space\space\space\blacksquare$
A: You are interested in showing that 
$$2^{n}+7 < 2^{n+1}=2(2^n)$$
which is equivalent to $7 < 2^n$ which is is true for $n \geq 6$.
Remark:
$\color{blue}{n < } 7n < 2^n $ and the two lines above the question mark seems redundant.
