Proof that $n^3+2n$ is divisible by $3$ I'm trying to freshen up for school in another month, and I'm struggling with the simplest of proofs!

Problem:
For any natural number $n , n^3 + 2n$ is divisible by $3.$
  This makes sense
Proof:
Basis Step: If $n = 0,$ then $n^3 + 2n = 0^3 +$
  $2 \times 0 = 0.$ So it is divisible by $3.$
Induction: Assume that for an arbitrary natural number $n$,
  $n^3+ 2n$ is divisible by $3.$ 
Induction Hypothesis: To prove this for $n+1,$ first try to express $( n + 1 )^3 + 2( n + 1 )$ in terms of $n^3 + 2n$ and use
  the induction hypothesis. Got it
$$( n + 1 )^3+ 2( n + 1 ) = (  n^3 + 3n^2+ 3n + 1 ) + ( 2n + 2 ) \{\text{Just some simplifying}\}$$
$$ = ( n^3 + 2n ) + ( 3n^2+ 3n + 3 ) \{\text{simplifying
 and regrouping}\}$$
  $$ = ( n^3 + 2n ) + 3( n^2 + n + 1 ) \{\text{factored out
the 3}\}$$
which is divisible by $3$, because $(n^3 + 2n )$ is divisible by $3$
  by the induction hypothesis. What?

Can someone explain that last part? I don't see how you can claim $(n^3+ 2n ) + 3( n^2 + n + 1 )$ is divisible by $3.$

The driving force behind induction is that you show that a base case (when $n = 1$, for example). Then you show that the hypothesis being true at some $k$ implies that it holds at $k+1$. Then, since you have verified the hypothesis at $n = 1$, you have it at $n = 2$. Then, since it holds at $n = 2$, it holds at $n = 3$, and so on. Note that the domain over which the hypothesis holds should be defined in the hypothesis itself.
Now, for your specific case, let's see how this works.
First, for the base case ($n = 1$), we see the following:
$$1^3 + 2 \cdot 1 = 3.$$
That is clearly divisible by $3$, so we have our base case.
Now assume that for all $k \geq 1$, $k^3 + 2k$ is divisible by $3$.
Then for $n = k + 1$, we have:
$$(k+1)^3 + 2(k+1) = k^3 + 3k^2 + 3k + 1 + 2k + 2 = k^3 + 3k^2 + 5k + 3$$
The $3k^2 + 3$ portion is clearly divisible by $3$, so we need only show that $k^3 + 5k$ is divisible by $3$. From the assumption above, we know that $k^3 + 2k = 3m$ for some positive integer $m$. Then,
$$k^3 + 5k = k^3 + 2k + 3k = 3m + 3k = 3(m + k),$$
so the hypothesis holds at $n = k+1$.
Thus, we have for all $n \geq 1$, $n^3 + 2n$ is divisible by $3$.
As the others have suggested, there are certainly other ways of showing the $(k+1)$th case, but hopefully this overall form helps you see how mathematical induction works.
