Prove by induction $2^n \geq n^3 \ \ \forall n\geq 10 $ Prove by induction $2^n \geq n^3 \ \ \forall n\geq 10 $
I did these steps:

*

*Basis step
$$P(10): \ \ 1024 \geq 1000 \ (True)$$

*Inductive step
$$P(n) \implies P(n+1) \\P(n+1) = 2^{n+1} \geq (n+1)^3$$
so $$2^n \geq n^3 \\ 2^n \cdot 2 \geq n^3  \cdot 2 \\ 2^{n+1} \geq 2n^3 \\ 2^{n+1} \geq (n+1)^3$$
taking advantage of the fact that $(n+1)^3 \geq 2n^3$ and less than $2^{n+1}$.

does this demonstration work?
I know there's a similar question but the solution is different, I want to know if my demonstration is valid as well.
 A: 
so $$2^n \geq n^3 \\ 2^n \cdot 2 \geq n^3  \cdot 2 \\ 2^{n+1} \geq 2n^3 \\ 2^{n+1} \geq (n+1)^3$$
taking advantage of the fact that $(n+1)^3 \geq 2n^3$ and less than $2^{n+1}$.


does this demonstration work?

No, your analysis (in and of itself) hasn't shown that 
$(n+1)^3 \leq 2^{n+1}$, assuming that I am not overlooking anything.  This is what you are trying to prove.
The easy way, is:
after noting that $2^{10} > 10^3$, and assuming that $2^n > n^3$, simply note that for $n \geq 10$,
$$\left(\frac{n+1}{n}\right)^3 < 2. \tag1$$
Note that $2^{1/3} > 1.25$, and for $n \geq 10, \frac{n+1}{n} < 1.25$.
This justifies equation (1) above.
Thus:
$$2^{n+1} = 2 \times 2^n > 2 \times n^3 > \left(\frac{n+1}{n}\right)^3 \times n^3 = (n+1)^3.$$
A: You already proved the base case
Now suppose that $n^3\le 2^n$  is true and
let's prove it for $n+1$
$(n+1)^3=n^3+\left(3n^2+3n+1\right)\tag{1}<\ldots$
for $n\ge 4$ we have $3n^2+3n+1 < n^3$ indeed, adding $n^3$ to both sides we get
$n^3+3 n^2+3 n+1<n^3+n^3\to (n+1)^3<2n^3\to n+1<n\sqrt[3]{2}\to n(\sqrt[3]{2}-1)>1\to n>\frac{1}{\sqrt[3]{2}-1}\approx 3.8$
continue $(1)$ $$\ldots<n^3+n^3\le 2^n+2^n=2^{n+1}$$
