Prove $\forall n \geq 10, 2^n > n^3$ Prove $\forall n \geq 10, 2^n > n^3$
base case: $n = 10$
$2^{10} = 1024$
$10^3 = 1000$
$1024 > 1024$.
So $P(k)$ holds for $k = n$. We seek to show $P(k+1)$ holds:

We know $2^k > k^3$.
$2^k+3k^2+3k+1>k^3+3k^2+3k+1 = (k+1)^3$
$\iff 2^k+3k^2+3k+1>(k+1)^3$
Let us compare $k^3$ and $3k^2+3k+1$:
$k^3$ $\Box$ $3k^2+3k+1$
$k^3-3k^2-3k$ $\Box$ $1$
$k(k^2-3k-3)$ $\Box$ $1$.
$k^2-3k-3 > 0$, $\forall k \geq 10$,
$\implies k^3 > 3k^2+3k+1, \forall k \geq 10$.
Therefore $2^k + k^3 > 2^k + (3k^2+3k+1)$.
We know $2^k > k^3$,
so $2^{k+1} = 2^k+2^k > 2^k + k^3 > 2^k+3k^2+3k+1 > (k+1)^3$
$\implies 2^{k+1} > (k+1)^3$.
Therefore $P(k+1)$ holds and our hypothesis is proven.

I don't really know how else to segue into the argument about comparing $k^3$ to $3k^2+3k+1$, is this ok notation?

I see your logic: you prove the chain
$$
2^{k+1}=2^k+2^k>k^3+k^3>k^3+3k^2+3k+1=(k+1)^3
$$
where the first inequality is due to the induction hypothesis and the second is because of the inequality $k^3>3k^2+3k+1$ ($k\geq 10$) for which you supply a separate proof. I think this is a good approach although you can write it more succinctly and clearly. You also can shorten the proof for $k^3>3k^2+3k+1$ as follows
$$
k\geq 10\implies k^3\geq10k^2>3k^2+3k^2+k^2>3k^2+3k+1.
$$
Alternatively, observe that
$$
2k^3>(k+1)^3\iff2>\left(1+\frac{1}{k}\right)^3\iff k>\frac{1}{\sqrt[3]{2}-1}\approx 3.84732
$$
so, of course, if $k\geq 10$ then you also have $2k^3>(k+1)^3$.
