<s>[INST] 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.
 [/INST] Hint: if $n\geq 10$, $$
\frac{(n+1)^3}{n^3}=\frac{n^3+3n^2+3n+1}{n^3}=1+\frac 3n+\frac 3{n^2}+\frac 1{n^3}\leq 1+\frac 3{10}+\frac 3{100}+\frac 1{1000}<2.
$$
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