Show that $0^i$ where $i$ is a power of 2 is not context free
If you're still confused, here's the proof. Let $p$ be the integer of the Pumping Lemma and consider the string $0^{2^p}$. If $L$ was a CFL, then (since $2^p\ge p$) the PL would apply to $0^{2^p}$ we'd have $0^{2^p}=uvxyz$ with


$|vxy|\le p$
$|vy| &gt; 0$


So if we let $v=0^k, y = 0^j$, we have from (1), $k + j =|vy| \le |vxy| \le p$ and from (2), $k + j=|vy|&gt;0$. In other words,


  $0&lt;k+j\le p$


Now pump up: The PL assures us that $uv^2xy^2z\in L$, and observe
$$
|uv^2xy^2z| = 2^p+(k+j)\le 2^p+p&lt; 2^p+2^p=2^{p+1}
$$
as long as $p&gt;0$, so the pumped string $uv^2xy^2z$ has length strictly between $2^p$ and $2^{p+1}$. No such string can be in $L$, so we have a contradiction to the PL implication that $uv^2xy^2z\in L$ and so our initial assumption, that $L$ was a CFL, must be false.

You could shorten this proof if you knew that (1) any context-free language over a 1-symbol alphabet must be regular and (2) the language $L=\{0^{2^n}\mid n\ge 0\}$ is not regular.