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?

Your proof is valid.  You segue is clunky but acceptable.  It's a matter of personal style but I'd do the following.
Induction step:  Assume for some $k$ that $2^k > k^3$.
If we could show that $k^3 > 3k^2 + 3k + 1$ we'd have:
$2^{k+1} = 2^k + 2^k > k^3 + k^3 > k^3 + 3k^2 + 3k + 1 = (k+1)^3$
and we'd be done.
So it suffices to prove $k^3 > 3k^2 + 3k + 1$ for $k \ge 10$.
So.....
But it's a matter of style.
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I'd also prove $k^3 > 3k^2 + 3k + 1$ via
$k^3 > 10k^2 = 3k^2 + 7k^2 > 3k^2 + 70k = 3k^2 + 3k + 67k > 3k^2 + 3k + 1$.
but your way works too.
Or... maybe I'd do it:
As $k  > 0$ I can conclude $k^3 > 3k^2 + 3k + 1$ when $k > 3 + 3/k + 1/k^2$. As $k > 3$ $3/k + 1/k^2 < 3/k + 1/k = 4/k \le 1$ if $k \ge 4$ so $k^3 > 3k^2 + 3k +1$ for all $k \ge 4$. 
