Consider, input alphabet $\Sigma = \{1\}$
So, set $L = \{1,11,1111,11111111,...\} = \{1^{2^n } \ | \ n \geq 0 \} $
You can prove the $L$ to be non-regular set by Pumping Lemma.
If $L$ is a regular set then
$\exists \ p$ such that $\forall z \in L, $ where $|z| \geq p$ ( $p$ is the number of states in the finite state machine for $L$, it is also called as pumping length.)
$\exists \ v,w,x$ such that $z = vwx$ and $|vw| \leq p \ ; |w| \geq 1$ and
$\forall \ i\geq 0, vw^ix \in L$
This is called the pumping property.
If $L$ is a regular set then it must satisfy the pumping property but it can be shown that it does not satisfy the pumping property by a contradiction.
Suppose, $L$ is a regular set.
Consider an element of this regular set as $z = 1^{2^{p-1}} 1^{2^{p-1}} \ $ with $p \geq 1$
Here, $z \in L$ and $|z| \geq p$
Now, try to decompose $z$ as $z = vwx = 1^a1^b11^{2^{p-1}-a-b-1} 1^{2^{p-1}} \ $ for $a,b \geq 0$ with
$v = 1^a , w= 1^b1, x = 1^{2^{p-1}-a-b-1} 1^{2^{p-1}}$
Here, $|vw| \leq p$ and $|w| \geq 1$
Now, pump $z$ two times i.e. for $i=2,$ $vw^ix$ is:
$vw^2x =1^a(1^b1)^2 1^{2^{p-1}-a-b-1} 1^{2^{p-1}}= 1^a 1^b11^b11^{2^{p-1}-a-b-1} 1^{2^{p-1}} = 1^{2^p +b+1}$
For $b=2, \ z = 1^{2^p+3} \notin L$
But according to pumping lemma, $z$ should be in $L.$ So, Contradiction.
Hence, $L$ is not a regular set.