Use induction to prove that $2^n \gt n^3$ for every integer $n \ge 10$. Use induction to prove that $2^n \gt n^3$ for every integer $n \ge 10$.
My method:
If $n = 10$, $2^n \gt n^3$ where $2^{10} \gt 10^3$ which is equivalent to $1024 \gt 1000$, which holds for $n = 10$. $2^k \gt k^3$. 
$2^{k + 1} \gt (k + 1)^3$
$2^{k + 1} \gt (k + 1)(k + 1)(k + 1)$
$2^{k + 1} \gt (k^2 + 2k + 1)(k + 1)$
$2^{k + 1} \gt (k^3 + 3k^2 + 3k + 1)$
$(k + 1) \cdot 2^k \gt k^3 + 3k^2 + 3k + 1$
Since $2^{10} \gt 10^3$, the inequality holds for $n = 10$. Assume $2^k \gt k^3$ for $k \ge 10$. We show that $2^{k + 1} \gt (k + 1)^3$. Hence, $2^{k + 1} = 2 \cdot 2^k \gt 2k^3$
$= k^3 + k^3 \ge k^3 + 10k^2$
$= k^3 + 10k^2 = k^3 + 4k^2 + 6k^2 \ge k^3 + 4k^2 + 6 \cdot 10 = k^3 + 4k^2 +60$
$k^3 + 4k^2 + 60 \gt k^3 + 3k^2 + 3k + 1$
Therefore, $2^{k + 1} \gt (k + 1)^3$. Hence, $2^n \gt n^3$ for every integer $n \ge 10$.
I was trying to fix this but I am not sure how to go about doing so.

We do the induction step, perhaps somewhat along the lines that you had considered. 
Let $k\ge 10$, and suppose that $2^k\gt k^3$. We want to show that $2^{k+1}\gt (k+1)^3$.
We have
$$2^{k+1}=2\cdot 2^k\gt k^3+k^3.$$
It will now be enough to show that $k^3+k^3\gt (k+1)^3$, or equivalently that $(k+1)^3\lt k^3+k^3$.
We have 
$$(k+1)^3=k^3+3k^2+3k+1\le k^3+3k^2+3k^2+k^2=k^3+7k^2.$$
But $7k^2\lt k^3$, and therefore $(k+1)^3\lt k^3+k^3$.
