Prove using Induction | Tricky one! I actually need some help. I want to prove using simple induction that
Q.1) $2^n > n^3$ for all $n \geq 10   $ 
I tried solving it like this...  

Base Step:

$n = 10$:
$2^{10} > 10^3 = 1024 > 1000$ So, that's true and fine.  

Inductive step:  

Suppose $2^n > n^3$ is true for some $n$. So it means that it should also be true for $n+1$ So,  
$2^{n+1} > (n+1)^3$
 $2^{n+1} > n^3 + 3n^2 + 3n + 1$   
L.H.S:
$2^{n+1} = $? 
Now here I'm stuck in further expanding this prove. I want to solve it further and make the $2^{n+1}$ with a power on top of it that I can use to compare with $n^3 + 3n^2 + 3n + 1$.  
Kindly, tell me as easy as possible as I'm not expert in it. Thanks   
 A: Hint:
From the induction hypothesis, you deduce that
$$2^{n+1}=2\cdot 2^n>2n^3,$$ 
hence by transitivity, it's enough to show that $2n^3\ge (n+1)^3$, or $\Bigl(1+\dfrac 1n\Bigr)^3\le2$.
Observe that
$$\Bigl(1+\dfrac 1n\Bigr)^3=1+\frac3n+\frac3{n^2}+\frac1{n^3}\le 1+\frac9n\quad\text{(why?)}$$
A: What you have to prove is that, assuming $2^n>n^3$, $2^{n+1}>(n+1)^3$.
As $2^{n+1}=2\times2^n>2n^3$ by hypothesis, it suffices to prove that $2n^3>(n+1)^3$ for $n\ge10$.
One way to do this is study function $x\mapsto x^3-3x^2-3x-1$ and prove it takes positive values for $x\ge10$.
A: You already verified that $10^3 \lt 2^{10}.$
For the induction step, if $n^3\lt 2^n$ and $n\ge 10,$ then
\begin{align}
(n+1)^3 &=\big(\frac{n+1}{n}\cdot n\big)^3
\\&=\big(1+\frac1{n}\big)^3 \cdot n^3
\\&\lt \big(1+\frac1{4}\big)^3 \cdot 2^n \scriptsize{\quad-\;\text{By (a) the fact that }n\gt 4\text{ and (b) the induction hypothesis.}}
\\&=\big( \frac5{4}\big)^3 \cdot 2^n
\\&=\frac{125}{64} \cdot 2^n
\\&\lt \frac{128}{64} \cdot 2^n
\\&= 2 \cdot 2^n
\\&= 2^{n+1}.
\end{align}
