# Use the Pumping Lemma to show $\Sigma^*\setminus\{0^n1^n: n\geq 0\}$ is not regular (without using complement closure)

Question: Use the Pumping Lemma to show $$L_1 = \Sigma^*\setminus\{0^n1^n: n\geq 0\}$$ is not regular, for $$\Sigma=\{0,1\}$$ (without using the complement closure property).

My thoughts: I understand that $$L_2 = \{0^n1^n: n\geq 0\}$$ can be shown to be not regular using the Pumping Lemma by starting with a string in $$L_2$$ of length at least the pumping length, and try to pump it outside the language.

So I've been trying, for $$L_1$$, to start with something and pump it to $$L_2$$, but I have been struggling to find a string such that all it's recompositions can be pumped into $$L_2$$. As it seems whether I can perform the pumping successfully always depends on the decomposition...

Suppose towards a contradiction that $$L_1$$ is regular and let $$p$$ be its pumping length.
Since $$0^{p}1^{p+p!} \in L$$ there must be some $$1 \le k \le p$$ such that $$0^{p+ik} 1^{p + p!} \in L_1$$ for every integer $$i \ge -1$$.
Choosing $$i=\frac{p!}{k}$$ (this is an integer since $$k$$ is a factor of $$p!$$) yields the following contradiction: $$0^{p+\frac{p!}{k} \cdot k} 1^{p + p!} = 0^{p + p!} 1^{p + p!} \in L_1.$$