Show $\{1^n0^m |\space n \neq 2^m\}$ not regular using pumping lemma Showing that the language $L$ with  $\{1^n0^m |\space n \neq 2^m\}$ is not regular using Myhill-Nerode is easy: Let $i, j\in \mathbb{N}.i\neq j.$ It follows $1^{2^i}\nsim 1^{2^j}$ because $1^{2^i}0^{i}\notin L$ but $1^{2^j}0^{i}\in L$. Therefore $L$ has an infinite amount of Myhill-Nerode equivalence classes and is not regular. 
But how do I show this using the general version of the pumping lemma for regular languages? https://en.wikipedia.org/wiki/Pumping_lemma_for_regular_languages#General_version_of_pumping_lemma_for_regular_languages
Let $p$ be the pumping length, and consider the string $u=1^{2^{p!+p}}$, $w = 0^p$, $v = \epsilon$. Notice $uwv \in L$. According to the pumping lemma, there is a value $q \in \{1,\ldots,p\}$ such that $1^{2^{p!+p}} 0^{p-q+iq} \in L$ for every $i \geq 0$. Choosing $i = 1 + p!/q$, we obtain a contradiction.