Direct Proof that $1 + 3 + 5 + \cdots+ (2n - 1) = n\cdot n$ 
Prove that for all integers $n$, $n \geq 1$,
  $$1 + 3 + 5 + \cdots + (2n - 1) = n\cdot n$$

How would I go about proving this?

By inspection note that $1 = 1 \cdot 1$, that $1 + 3 = 2 \cdot 2$, and that $1 + 3 + 5 = 3 \cdot 3$.
Next, witness that the difference between any two successive perfect squares $(n - 1)^2$ and $n^2$ is a particular odd integer:
$n^2 - (n - 1)^2 \implies n^2 - (n^2 - 2n + 1) \implies 2n - 1$
Note that this is the very formula which you are summing to generate your series.
So for instance the difference between $3 \cdot 3$ and $2 \cdot 2$ is $2 \cdot 3 - 1 = 5$, corresponding to the $5$ in $3\cdot3 = 1 + 3 + 5$.
If the sum of the series generated so far is a square (shown by inspection), and the next term is from the series generated by $2n - 1$, then after this term is added, the sum must be the next square because the each term obeys the formula for the difference between two successive squares.
