Prove that a language is not regular with process of elimination
A simple technique is using the pumping lemma:

For any regular language $L$ there exists an integer $n$ such that for all $x \in L$, with $|x| \geq n$ there exists $u,v,w \in \Sigma^*$, such that,


$x = uvw$
$|uv| \leq n$
$|v| \geq 1$
for all $i \geq 0: uv^iw \in L$


Assumption: Suppose your language is regular.

Consider $s = a^nb^n$. Clearly $s \in L$ and $|s| \geq n$. We get that $uvw = a^nb^n$ and we know that $|uv| \leq n$, i.e $v$ can only contain $a$'s and since $|v| \geq 1$ it must contain at least one $a$. According to the last condition of the pumping lemma, $uv^0w \in L$ but this is clearly not the case. We now have a contradiction as we assumed the language to be regular:

The assumption must be wrong implying non-regularity of the language