2 to the n is Greater than n Cubed when n is 10 and above

Theorem

 * $\forall n \in \Z, n \ge 10: 2^n > n^3$

Proof
Proof by induction:

For all $n \in \Z$ such that $n \ge 10$, let $P \left({n}\right)$ be the proposition:
 * $2^n > n^3$

We note that:
 * $2^9 = 512 < 729 = 9^3$

so when $n < 10$ the proposition does not hold.

Basis for the Induction
$P \left({10}\right)$ is the case:
 * $2^{10} = 1024 > 1000 = 10^3$

so $P \left({0}\right)$ holds.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 10$, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is our induction hypothesis:
 * $2^k > k^3$

We need to show that:
 * $2^{k + 1} > \left({k + 1}\right)^3$

Induction Step
This is our induction step:

We note that when $k \ge 10$, we have:
 * $(1): \quad \left({1 + \dfrac 1 k}\right)^3 < 2$

Thus:

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.