Is the language $L=\{a^{2^{n}} \mid$ n is a natural number$\} $ context free?
Suppose that $L$ is CFL then $L$ is regular because it is a unary language. So exist a constant $N$ associated to $L$ by the pumping lemma for regular languages.

We choose the word $\sigma = a^{2^N}$ and $|\sigma| = 2^N &gt; N $ so the pumping lemma should complies for $\sigma$. Then exist a factorization of $\sigma = \alpha \beta \gamma$ such that:

1) $|\alpha \beta|\leq N $

2) $|\beta|\geq 1$

3) $\alpha\beta^i\gamma \in L \forall i \in \mathbb{N}_0$

$|\alpha\beta^i\gamma|= 2^N + |\beta|(i-1)$ choosing $i=2$:

$|\alpha\beta^2\gamma|= 2^N + |\beta|$ so:
$$2^N+1\leq|\alpha\beta^2\gamma|\leq 2^N+N$$
$$2^N&lt;2^N+1\leq|\alpha\beta^2\gamma|\leq 2^N+N&lt;2^N +2^N$$
$$2^N&lt;|\alpha\beta^2\gamma|&lt;2^{N+1}$$

Then $|\alpha\beta^2\gamma|$ can't be a power of 2(with natural exponent) so $\alpha\beta^2\gamma\notin L$ and we find a contradiction. Finally $L$ can't be regular and it is not CFL too.