Is $2^{2n} = O(2^n)$? Is $2^{2n} = O(2^n)$?
My solution is:
$2^n 2^n \leq C_{1}2^n$
$2^n \leq C_{1}$,
TRUE.
Is this correct?

$x^n=o(y^n)$ iff $x&lty$, as $(\frac{x}{y})^n\rightarrow 0$. Here 4 and 2, so no.
