<s>[INST] 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?
 [/INST] $$\begin{eqnarray*} (1 + 3 + 5 + ... + (2n - 1)) &=& (1 + 2 + 3 + ... + 2n) - (2 + 4 + ... + 2n)\\
&=& \sum_{i=1}^{2n}{i} - 2\sum_{i=1}^{n}{i}\\
&=& \frac{(2n)(2n+1)}{2} - 2\left(\frac{n(n+1)}{2}\right)\\
&=& n(2n+1) - n(n+1)\\
&=& n(2n + 1 - n - 1) = n^{2}\end{eqnarray*}$$
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